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4.2: The Trigonometric Series - Mathematics


4.2.1 Periodic functions and motivation

As motivation for studying Fourier series, suppose we have the problem

[ x'' + omega^2_0 x=f(t), ]

for some periodic function ( f(t)). We have already solved

[ x''+ omega^2_0 x=F_0 cos( omega t). ]

One way to solve (4.2.1) is to decompose (f(t)) as a sum of cosines (and sines) and then solve many problems of the form (4.2.2). We then use the principle of superposition, to sum up all the solutions we got to get a solution to (4.2.1).

Before we proceed, let us talk a little bit more in detail about periodic functions. A function is said to be periodic with period (P) if ( f(t)) for all (t). For brevity we will say ( f(t)) is (P-)periodic. Note that a (P-)periodic function is also (2P-)periodic, (3P-)periodic and so on. For example, ( cos(t)) and ( sin(t)) are ( 2 pi -)periodic. So are ( cos(kt)) and ( sin(kt)) for all integers ( k). The constant functions are an extreme example. They are periodic for any period (exercise).

Normally we will start with a function (f(t)) defined on some interval ( [-L, L]) and we will want to extend periodically to make it a ( 2L-)periodic function. We do this extension by defining a new function (F(t)) such that for (t) in( [-L, L]), ( F(t)=f(t)). For (t) in ( [L, 3L]), we define ( F(t)=f(t-2L)), for (t) in ( [-3L, -L]), ( F(t)=f(t+2L)), and so on. We assumed that ( f(-L)=f(L)). We could have also started with (f) defined only on the half-open interval ( (-L, L]) and then define ( f(-L)=f(L)).

Example (PageIndex{1}):

Define ( f(t)=1-t^2) on ([-1, 1]). Now extend periodically to a 2-periodic function. See Figure 4.2.

Figure 4.2: Periodic extension of the function ( 1-t^2).

You should be careful to distinguish between ( f(t)) and its extension. A common mistake is to assume that a formula for ( f(t)) holds for its extension. It can be confusing when the formula for ( f(t)) is periodic, but with perhaps a different period.

Exercise (PageIndex{1}):

Define ( f(t)= cos t) on ([dfrac{ - pi}{2}, dfrac{ pi}{2} ]). Take the (pi -)periodic extension and sketch its graph. How does it compare to the graph of ( cos t)?

4.2.2 Inner product and eigenvector decomposition

Suppose we have a symmetric matrix, that is ( A^T=A). We have said before that the eigenvectors of ( A) are then orthogonal. Here the word orthogonal means that if ( vec{v}) and ( vec{w}) are two distinct (and not multiples of each other) eigenvectors of ( A), then ( left langle vec{v}, vec{w} ight angle=0). In this case the inner product (left langle vec{v}, vec{w} ight angle) is the dot product, which can be computed as ( vec{v}^T vec{w}).

To decompose a vector ( vec{v}) in terms of mutually orthogonal vectors ( vec{w}_1) and ( vec{w}_2) we write

[ vec{v} = a_1 vec{w}_1+a_2 vec{w}_2.]

Let us find the formula for (a_1) and (a_2). First let us compute

[ left langle vec{v}, vec{w}_1 ight angle = left langle a_1 vec{w}_1 + a_2 vec{w}_2, vec{w}_1 ight angle = a_1 left langle vec{w}_1, vec{w}_1 ight angle + a_2 left langle vec{w}_2, vec{w}_1 ight angle= a_1 left langle vec{w}_1, vec{w}_1 ight angle. ]

Therefore,

[ a_1= dfrac{left langle vec{v}, vec{w}_1 ight angle}{left langle vec{w}_1, vec{w}_1 ight angle}.]

Similarly

[ a_2= dfrac{left langle vec{v}, vec{w}_2 ight angle}{left langle vec{w}_2, vec{w}_2 ight angle}.]

You probably remember this formula from vector calculus.

Example (PageIndex{2})

Write ( vec{v}=left[ egin{array}{c} 2 3 end{array} ight] ) as a linear combination of ( vec{w}_1=left[ egin{array}{c} 1 -1 end{array} ight] ) and ( vec{w}_2=left[ egin{array}{c} 1 1 end{array} ight] ).

First note that ( vec{w}_1) and ( vec{w}_2) are orthogonal as ( left langle vec{w}_1, vec{w}_2 ight angle = 1(1)+(-1)1=0). Then

[ a_1= dfrac{left langle vec{v}, vec{w}_1 ight angle}{left langle vec{w}_1, vec{w}_1 ight angle}= dfrac{2(1)+3(-1)}{1(1)+(-1)(-1)}=dfrac{-1}{2}, a_2= dfrac{left langle vec{v}, vec{w}_2 ight angle}{left langle vec{w}_2, vec{w}_2 ight angle}=dfrac{2+3}{1+1}= dfrac{5}{2}.]

Hence

[ left[ egin{array}{c} 2 3 end{array} ight]=dfrac{-1}{2} left[ egin{array}{c} 1 -1 end{array} ight]+ dfrac{5}{2} left[ egin{array}{c} 1 1 end{array} ight]. ]

4.2.3 The Trigonometric Series

Instead of decomposing a vector in terms of eigenvectors of a matrix, we will decompose a function in terms of eigenfunctions of a certain eigenvalue problem. The eigenvalue problem we will use for the Fourier series is

[ x'' + lambda x=0,~~~~ x(- pi)=x(pi)~~~~x'(- pi)=x'( pi).]

We have previously computed that the eigenfunctions are (1, cos(kt), sin(kt)). That is, we will want to find a representation of a ( 2 pi -)periodic function ( f(t)) as

[ f(t)= dfrac{a_0}{2}+ sum^{infty}_{n=1}a_n cos(nt)+b_n sin(nt).]

This series is called the Fourier series2 or the trigonometric series for (f(t)). We write the coefficient of the eigenfunction (1) as ( dfrac{a_0}{2}) for convenience. We could also think of ( 1= cos(0t)), so that we only need to look at ( cos(kt)) and ( sin(kt)).

As for matrices we want to find a projection of (f(t)) onto the subspace generated by the eigenfunctions. So we will want to define an inner product of functions. For example, to find ( a_n) we want to compute ( left langle f(t), cos(nt) ight angle ). We define the inner product as

[ left langle f(t), g(t) ight angle = int^{pi}_{-pi} f(t)g(t)dt. ]

With this definition of the inner product, we have seen in the previous section that the eigenfunctions ( cos(kt))(including the constant eigenfunction), and ( sin(kt)) are orthogonal in the sense that

[ left langle cos(mt), cos(nt) ight angle = 0 ~~~~ { m{for}}~ m eq n, left langle sin(mt), sin(nt) ight angle = 0 ~~~~ { m{for}}~ m eq n, left langle sin(mt), cos(nt) ight angle = 0 ~~~~ { m{for~all}}~ m { m{~and~}} n.]

By elementary calculus for ( n=1,2,3, ldots .) we have ( left langle cos(nt), cos(nt) ight angle = pi) and ( left langle sin(nt), sin(nt) ight angle = pi). For the constant we get that ( left langle 1, 1 ight angle = 2 pi). The coefficients are given by

[ a_n= dfrac{ left langle f(t), cos(nt) ight angle}{left langle cos(nt), cos(nt) ight angle}= dfrac{1}{ pi} int^{pi}_{-pi} f(t) cos(nt)dt, b_n= dfrac{ left langle f(t), sin(nt) ight angle}{left langle sin(nt), sin(nt) ight angle}= dfrac{1}{ pi} int^{pi}_{-pi} f(t) sin(nt)dt. ]

Compare these expressions with the finite-dimensional example. For ( a_0) we get a similar formula

[ a_0 = 2 dfrac{ left langle f(t), 1 ight angle}{left langle 1, 1 ight angle} dfrac{1}{ pi} int^{pi}_{-pi} f(t)dt. ]

Let us check the formulas using the orthogonality properties. Suppose for a moment that

[ f(t)= frac{a_0}{2}+ sum^infty_{n=1}a_n cos(nt)+b_n sin(nt). ]

Then for ( m geq 1) we have

[ left langle f(t), cos(mt) ight angle = left langle frac{a_0}{2}+ sum^infty_{n=1}a_n cos(nt)+b_n sin(nt), cos(mt) ight angle =frac{a_0}{2}left langle 1, cos(mt) ight angle +sum^infty_{n=1}a_n left langle cos(nt), cos(mt) ight angle + b_n left langle sin(nt), sin(mt) ight angle = a_m left langle cos(mt), cos(mt) ight angle . ]

And hence ( a_m=frac{left langle f(t), cos(mt) ight angle}{left langle cos(mt), cos(mt) ight angle}.)

Exercise (PageIndex{2}):

Carry out the calculation for (a_0) and (b_m).

Example (PageIndex{3}):

Take the function

[ f(t)=t]

for (t) in ((- pi, pi]). Extend ( f(t)) periodically and write it as a Fourier series. This function is called the sawtooth.

Figure 4.3: The graph of the sawtooth function.

The plot of the extended periodic function is given in Figure 4.3. Let us compute the coefficients.

Solution

We start with (a_0),

[ a_0 = frac{1}{pi} int^pi_{-pi} tdt=0.]

We will often use the result from calculus that says that the integral of an odd function over a symmetric interval is zero. Recall that an odd function is a function ( varphi(t)) such that ( varphi(-t) = - varphi(t)). For example the functions ( t, sin t), or (importantly for us) ( t cos(nt)) are all odd functions. Thus

[ a_n=frac{1}{pi} int^pi_{-pi} t cos(nt)dt=0.]

Let us move to ( b_n). Another useful fact from calculus is that the integral of an even function over a symmetric interval is twice the integral of the same function over half the interval. Recall an even function is a function (varphi(t)) such that ( varphi(-t) = varphi(t)). For example ( t sin(nt)) is even.

[ b_n= frac{1}{pi} int^pi_{-pi} t sin(nt)dt = frac{2}{pi} int^pi_{0} t sin(nt)dt = frac{2}{pi} left( left[ frac{-t cos(nt)}{n} ight] ^pi_{t=0}+frac{1}{n} int^pi_{0} cos(nt)dt ight) = frac{2}{pi} left( frac{- pi cos(n pi)}{n}+0 ight) = frac{-2 cos(n pi)}{n}=frac{2(-1)^{n+1}}{n}.]

We have used the fact that

[ cos(n pi)=(-1)^n= left{ egin{array}{c} 1~~~~ { m{~if~}} n { m{~even,~}} -1~~~~ { m{~if~}} n { m{~odd.~}} end{array} ight.]

The series, therefore, is

[ sum^infty_{n=1} frac{2(-1)^{n+1}}{n} sin(nt).]

Let us write out the first 3 harmonics of the series for (f(t)).

[ 2 sin(t)- sin(2t)+ frac{2}{3} sin(3t)+ cdots]

The plot of these first three terms of the series, along with a plot of the first 20 terms is given in Figure 4.4.

Figure 4.4: First 3 (left graph) and 20 (right graph) harmonics of the sawtooth function.

Example (PageIndex{4}):

Take the function

[ f(t)= left{ egin{array}{cc} 0&~~~~ { m{~if~}} - pi < t leq 0, pi& { m{~if~}} 0 < t leq pi. end{array} ight.]

Extend (f(t)) periodically and write it as a Fourier series. This function or its variants appear often in applications and the function is called the square wave.

Figure 4.5: The graph of the square wave function.

The plot of the extended periodic function is given in Figure 4.5. Now we compute the coefficients. Let us start with (a_0)

[ a_0= frac{1}{pi} int^pi_{-pi} f(t)dt= frac{1}{pi} int^pi_{0} pi dt= pi . ]

Next,

[ a_n = frac{1}{pi} int^pi_{-pi} f(t) cos(nt)dt= frac{1}{pi} int^pi_{0} pi cos(nt)dt= 0 . ]

And finally

[b_n = frac{1}{pi} int^pi_{-pi} f(t) sin(nt)dt = frac{1}{pi} int^pi_{0} pi sin(nt)dt = left[ frac{- cos(nt)}{n} ight]_{t=0}^{pi} =frac{1- cos( pi n)}{n}=frac{1-(-1)^n}{n}= left{ egin{array}{c} frac{2}{n}~~~~ { m{~if~}} n { m{~is~odd,~}} 0~~~~ { m{~if~}} n { m{~is~even.~}} end{array} ight.]

The Fourier series is

[ frac{pi}{2} + sum^{infty}_{n=1~ n~ m{odd}} frac{2}{n} sin(nt)+sum^{infty}_{k=1} frac{2}{2k-1} sin((2k-1)t).]

Let us write out the first 3 harmonics of the series for (f(t)).

[ frac{pi}{2}+2 sin(t)+ frac{2}{3} sin(3t)+ cdots]

The plot of these first three and also of the first 20 terms of the series is given in Figure 4.6.

Figure 4.6: First 3 (left graph) and 20 (right graph) harmonics of the square wave function.

We have so far skirted the issue of convergence. For example, if (f(t)) is the square wave function, the equation

[ f(t)= frac{pi}{2} + sum_{k=1}^{infty}frac{2}{2k-1} sin((2k-1)t).]

is only an equality for such (t) where (f(t)) is continuous. That is, we do not get an equality for (t= - pi, 0, pi) and all the other discontinuities of (f(t)). It is not hard to see that when (t) is an integer multiple of (pi) (which includes all the discontinuities), then

[ frac{pi}{2} + sum_{k=1}^{infty}frac{2}{2k-1} sin((2k-1)t)=frac{pi}{2}.]

We redefine (f(t)) on ([- pi, pi]) as

[ f(t)=left{ egin{array}{cc} 0 &~~~~ { m{~if~}} - pi

and extend periodically. The series equals this extended (f(t)) everywhere, including the discontinuities. We will generally not worry about changing the function values at several (finitely many) points.

We will say more about convergence in the next section. Let us however mention briefly an effect of the discontinuity. Let us zoom in near the discontinuity in the square wave. Further, let us plot the first 100 harmonics, see Figure 4.7. You will notice that while the series is a very good approximation away from the discontinuities, the error (the overshoot) near the discontinuity at ( t= pi) does not seem to be getting any smaller. This behavior is known as the Gibbs phenomenon. The region where the error is large does get smaller, however, the more terms in the series we take.

Figure 4.7: Gibbs phenomenon in action.

We can think of a periodic function as a “signal” being a superposition of many signals of pure frequency. For example, we could think of the square wave as a tone of certain base frequency. It will be, in fact, a superposition of many different pure tones of frequencies that are multiples of the base frequency. On the other hand a simple sine wave is only the pure tone. The simplest way to make sound using a computer is the square wave, and the sound is very different from nice pure tones. If you have played video games from the 1980s or so, then you have heard what square waves sound like.


Simplify to one trigonometric expression: 2 sin (pi/4) 2 cos (pi/4)

Got stuck on this one even after looking at a solution online:

Looks familiar, like the double angle sin

sin (( 2(pi/4 ) = 2 sin (pi/4) cos (pi/4)

so sin (pi/2) = 2 sin (pi/4) cos (pi/4)

the last step shown says that therefore

2 sin (pi/2) = 2 sin (pi/4) 2 cos (pi/4)

So the dark green part is part of the double angle formula

2 sin θ = cos θ - sin(2θ) putting that back into. bleh, my brain is fried today. I think factoring out 1/2 comes next somehow? Any help would be appreciated.

Dr.Peterson

Elite Member

Got stuck on this one even after looking at a solution online:

Looks familiar, like the double angle sin

sin (( 2(pi/4 ) = 2 sin (pi/4) cos (pi/4)

so sin (pi/2) = 2 sin (pi/4) cos (pi/4)

the last step shown says that therefore

2 sin (pi/2) = 2 sin (pi/4) 2 cos (pi/4)

So the dark green part is part of the double angle formula

2 sin θ = cos θ - sin(2θ) putting that back into. bleh, my brain is fried today. I think factoring out 1/2 comes next somehow? Any help would be appreciated.

The green part is the answer. Essentially, you have doubled the double angle formula to get 2 sin(pi/2) = 2 sin(pi/4) 2 cos(pi/4).

But why did you then subtract cosθ? That would be valid if it were being added, but you don't undo multiplication by subtracting.

Furthermore, there is no need to undo anything! What you started with was the right side of the equation above you have shown that it is equal to the left side, so the left side is the simplified answer: 2 sin(pi/2). What answer did this site give? Wasn't it that?

And if you haven't actually stated the original problem you are working on, can you quote it exactly for us? I have to say, "one trigonometric expression" sounds like there should be no coefficient on the outside of the answer, which can't happen here as far as I can see and the 2 in the middle seems odd for an expression as given.

Math_knight

New member

Hey Dr. Peterson, thank you for your reply. Sorry about any confusion. To sort of start over the problem gives us the expression "2 sin (2x) = 2 sin (x) 2 cos (x)" and then asks us to "simplify to one trigonometric expression"

So to your point, I guess they meant go from have two trig expressions (2sinxCosx) to one, and that it's ok to have a coefficient.

And the problem is an odd numbered one so the solution in the back of the book is indeed

The solution online shows most of the work, but not all of it because of course the person solving it takes it for granted that you know what he's thinking.

Anyway my question is about what you said " Essentially, you have doubled the double angle formula". How did they get the 2 in front of the sin (pi/2) AND the 2 in front of the cos

See, I only follow the solution as to this point:

sin (pi/2) = 2 sin (pi/4) cos (pi/4)

So far there is no 2 coefficient on the left side, NOR before the cos. I see that this is a a double angle formula with x= pi/4, so the double angle became "hidden" on the left side. Pi/2 is (2(pi/4)

The next step just makes the 2 coefficient appear on the left side and on the right side before the cos. Please note the orange "2"s. Poof.

2 sin (pi/2) = 2 sin (pi/4) 2 cos (pi/4)

So multiplying the left side by 2. multiplies the cos on the right side (and not the sin) by 2?

2 sin (2x) = 2 sin (x) 2 cos (x)

Why does multiplying the left side by 2 make a 2 appear in front of the cos on the right side? Best I can guess is that it's a transform? 2 sin x would double the amplitude? And. I have no idea..I'm just guessing. I have no idea how the cos x on the right side is related to an amplitude transform.

edit: it's probably somewhere in the proof of the double angle formula? which I will go over

edit #2: what would tripling the double angle formula do? this? 3 sin (2x) = 2 sin (x) 3 cos (x) ?


Snapshots


Integration by Trigonometric Substitution

Depending on the function we need to integrate, we can use this trigonometric expression as substitution to simplify the integral:

Example 1: Find the integral $ displaystyle int<><^<2>>+16>>>>>>$?

If we rewrite the expression inside the integral as $ displaystyle sqrt<<<^<2>>+16>>=sqrt<<<<<(4)>>^<2>>+<^<2>>>>$ then we see that we have the second form above that is:

So we will substitute $ displaystyle x=4 an heta $ and have $ displaystyle dx=4<^<2>> heta d heta $

The helpful trigonometric identities we will use is $ displaystyle <^<2>> heta =1+<< an >^<2>> heta $

$ displaystyle sqrt<<16<<>^<2>> heta >>=4sec heta $

Substituting $ displaystyle 4sec heta $ and $ displaystyle dx=4<^<2>> heta d heta $ into the given integral:

Now we write the answer in terms of x.

$ displaystyle an heta =frac<4>$

Substituting, we get the final answer:

Example 2: Find the integral $ displaystyle int<^<2>>-9>>>>dx$?

If we rewrite the expression inside the integral as $ displaystyle sqrt<<4<^<2>>-9>>=sqrt<<<<<(2x)>>^<2>>-<<<(3)>>^<2>>>>$ then we see that we have the third form above that is:

So we will$ displaystyle x=frac<3><2>sec heta $ and $ displaystyle dx=frac<3><2>sec heta an heta d heta $

The helpful trigonometric identities we will use$ displaystyle << an >^<2>> heta =<^<2>> heta -1$

Then we will have $ displaystyle sqrt<<4<^<2>>-9>>=sqrt<<4<<<(frac<3><2>sec heta )>>^<2>>-9>>$

$ displaystyle sqrt<<<<< an >>^<2>> heta >>= an heta $

Substituting $ displaystyle x=frac<3><2>sec heta $ and $ displaystyle dx=frac<3><2>sec heta an heta d heta $ into the given integral:

$ displaystyle frac<3><2>int<<<<< an >>^<2>>>> heta sec heta d heta =frac<3><2>int<<(<<>^<2>> heta -1)>>sec heta d heta =$

$ displaystyle frac<3><4>(sec heta an heta -ln (sec heta + an heta )+C$

Now we write the answer in terms of x.

$ displaystyle sec heta =frac<<2x>><3>$ or ($ displaystyle cos heta =frac<3><<2x>>$)


Sequence and Series – Practice Questions

Q1. Suppose α , β are the roots of x 2 + k x + 1 = 0 and γ , δ are the roots of x 2 + x + k = 0 .

(a) if α , β , γ , δ are in arithmetic sequence, then find the common difference of this arithmetic sequence in terms of k ,

(b) if α , β , γ , δ are in geometric sequence, then find the common ratio of this geometric sequence in terms of k .

Q2. If the third term of a geometric sequence is 3 , then find the product of the first 5 terms?

Q3. The sum of an infinite geometric series with first term x is equal to 3. Show that – 6 < x < 0 .

Q4. Find the coefficient of x 49 in the polynomial p ( x ) = ( x – 1 ) ( x – 2 ) ( x – 3 ) . . . . . . . . . . . . ( x – 50 ) .

Q5. If S 1 , S 2 , S 3 , . . . . . . . . . . . . , S n are the sum of infinite geometric series whose first terms are 2 , 3 , 4 , . . . . . . . . . . ( n – 1 ) and whose common ratios are

1 3 , 1 4 , 1 5 , . . . . . . . . . . , 1 n respectively then find the value of S 1 + S 2 + S 3 + . . . . . . . . . . . . . + S 2 n .

Q6. Find in a terms of c if 1 , log b a , log c b , – 24 log a c are in arithmetic sequence.

Q7. Find the value of x if ln ( 3 x – 1 ) , ln ( 3 x + 1 ) , ln ( 3 x + 1 – 1 ) are in arithmetic sequence.

Q8. Find the sum to n terms of the series: 3 2 + 5 4 + 9 8 + 17 16 + . . . . . . . . . . .

Q9. ln a – ln 2 b , ln 2 b – ln 3 c , ln 3 c – ln a are in arithmetic sequence. If a , b , c

are in geometric sequence, show that a = 9 c 4 .

Q10. Two arithmetic sequences have the sum of their n

terms in the ratio 3 n : ( 2 n – 1 ) . Find the ratio of their 23 rd terms.

Q11. Show that the sum to n terms of the series 2 + 8 3 + 26 9 + 80 27 + 242 81 + . . . . . . . . . is 3 n – 3 2 + 1 2 × 3 n – 1 .

Q12. If S = 1 4 + 1 4 2 + 1 4 3 + . . . . . . . . . . . . ∞ , then find the value of 0 . 5 log 3 S .

Q13. Find the values of a , b and c if

x 11 – 1 x – 1 = 1 + a x + b x 2 + c x 3 + . . . . . . . . . . . . + x 10 where a , b , c ∈ ℝ .

Q14. If a , b and c are the m t h , n t h and p t h terms respectively of an arithmetic sequence and also a geometric sequence, then show that the value of a b – c b c – a c a – b is equal to 1.

Q15. If x , y , z are respectively the p t h , q t h , r t h of a geometric sequence, then show that the value of q – r ln x + r – p ln y + p – q ln z is equal to zero.

Q16. (a) A geometric progression has a second term of 12 and a sum to infinity of 54. Find the possible values of the first term of the progression. [4 marks]

(b) The n th term of a progression is p + q n , where p and q are constants, and S n is the sum of the first n terms.
(i) Find an expression, in terms of p , q and n , for S n . [3 marks]

(ii) Given that S 4 = 40 and S 6 = 72 , find the values of p and q . [2 marks]

Q17. A sequence a 1 , a 2 , a 3 , . . . . . is defined by

a 1 = 1 a n + 1 = k a n + 1 a n , n ≥ 1

where k is a positive constant.
(a) Write down expressions for a 2 and a 3 in terms of k , giving your answers in their simplest form. (3 marks)

Given that ∑ r = 1 3 a r = 10
(b) find an exact value for k . (3 marks)

Q18. A company, which is making 140 bicycles each week, plans to increase its production.
The number of bicycles produced is to be increased by d each week, starting from 140 in week 1, to 140 + d in week 2, to 140 + 2d in week 3 and so on, until the company is producing 206 in week 12.
(a) Find the value of d. (2 marks)
After week 12 the company plans to continue making 206 bicycles each week.
(b) Find the total number of bicycles that would be made in the first 52 weeks starting from and including week 1. (5 marks)

Q19. The first three terms of a geometric sequence are 7 k – 5 , 5 k – 7 , 2 k + 10 where k is a constant.
(a) Show that 11 k 2 – 130 k + 99 = 0 (4 marks)
Given that k is not an integer,
(b) show that k = 9 11 (2 marks)
For this value of k ,
(c) (i) evaluate the fourth term of the sequence, giving your answer as an exact fraction,
(ii) evaluate the sum of the first ten terms of the sequence. (6 marks)

Q20. A sequence u 1 , u 2 , u 3 , . . . . . . . is defined by u 1 = 7 , u n + 1 = u n + 15 .
The sum of the first n terms of this sequence is denoted by S n . The terms of a second sequence v 1 , v 2 , v 3 , . . . . . . form a geometric progression with first term 1.2 and common ratio 1.2.
(i) Show that u 3 + v 3 = 38 . 728 . [2 marks]
(ii) Show that S 70 = 36 715 . [3 marks]
(iii) Find the largest value of p such that v p < S 70 . [3 marks]
(iv) Find the largest value of q such that S q < v 70 . [4 marks]

Q21. The first three terms of a sequence are given by 5 x + 8 , – 2 x + 1 , x – 4
(a) When x = 11 , show that the first three terms form the start of a geometric sequence, and state the value of the common ratio.

(b) Given that the entire sequence is geometric for x = 11
(i) state why the associated series has a sum to infinity
(ii) calculate this sum to infinity.

(c) There is a second value for x that also gives a geometric sequence.
For this second sequence
(i) show that x 2 – 8 x – 33 = 0
(ii) find the first three terms
(iii) state the value of S 2 n and justify your answer.

22. A geometric sequence has first term 80 and common ratio 1 3 .
(a) For this sequence, calculate:
(i) the 7 t h term [2 marks]
(ii) the sum to infinity of the associated geometric series. [2 marks]
The first term of this geometric sequence is equal to the first term of an arithmetic sequence.
The sum of the first five terms of this arithmetic sequence is 240.
(b) (i) Find the common difference of this sequence. [2 marks]
(ii) Write down and simplify an expression for the nth term. [1 mark]
Let S n represent the sum of the first n terms of this arithmetic sequence.
(c) Find the values of n for which S n = 144 . [3 marks]

23. (a) Three consecutive terms in an arithmetic sequence are 3 e – p , 5 , 3 e p . Find the possible values of p . Give your answers in an exact form.

(b) Prove that there is no possible value of q for which 3 e – q , 5 , 3 e q are consecutive terms of a geometric sequence.


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4.2: The Trigonometric Series - Mathematics


Since $ displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 longrightarrow $ $ (adjacent)^2 + (x)^2 = (2)^2 longrightarrow adjacent = sqrt <4-x^2> longrightarrow $ $ cos heta = displaystyle< adjacent over hypotenuse >= displaystyle < sqrt<4-x^2>over 2 > $ Then $ displaystyle < 2 heta + 2 sin heta cos heta >+ C = 2 arcsin Big( frac <2>Big) + 2 cdot displaystyle < x over 2>cdot displaystyle < sqrt<4-x^2>over 2> $ $ = displaystyle 2 arcsin Big( frac <2>Big) + frac<1> <2>x cdot sqrt <4-x^2>+ C $ When using the method of trig substitution, we will always use one of the following three well-known trig identities :

        • (I) $ 1 - sin^2 heta = cos^2 heta $
        • (II) $ 1 + an^2 heta = sec^2 heta $ and
        • (III) $ sec^2 heta - 1 = an^2 heta $
        • 1.) $ displaystyle < int cos x , dx > = sin x + C $
        • 2.) $ displaystyle < int sin x , dx > = - cos x + C $
        • 3.) $ displaystyle < int sec^2 x , dx > = an x + C $
        • 4.) $ displaystyle < int csc^2 x , dx > = - cot x + C $
        • 5.) $ displaystyle < int sec x an x , dx > = sec x + C $
        • 6.) $ displaystyle < int csc x cot x , dx > = - csc x + C $
        • 7.) $ displaystyle < int an x , dx > = ln | sec x | + C $
        • 8.) $ displaystyle < int cot x , dx > = ln | sin x | + C $
        • 9.) $ displaystyle < int sec x , dx > = ln | sec x + an x| + C $
        • 10.) $ displaystyle < int csc x , dx > = ln | csc x - cot x | + C $

            • A.) $ sin 2x = 2 sin x cos x $
            • B.) $ cos 2x = 2 cos^2 x - 1 $ so that $ cos^2 x = displaystyle< frac<1><2>(1+cos 2x)> $
            • C.) $ cos 2x = 1 - 2 sin^2 x $ so that $ sin^2 x = displaystyle< frac<1><2>(1-cos 2x) > $
            • D.) $ cos 2x = cos^2 x - sin^2 x $

            Click HERE to see a detailed solution to problem 1.

            Click HERE to see a detailed solution to problem 2.

            Click HERE to see a detailed solution to problem 3.

            Click HERE to see a detailed solution to problem 4.

            Click HERE to see a detailed solution to problem 5.

            Click HERE to see a detailed solution to problem 6.

            Click HERE to see a detailed solution to problem 7.

            Click HERE to see a detailed solution to problem 8.

            Click HERE to see a detailed solution to problem 9.

            Click HERE to see a detailed solution to problem 10.

            Click HERE to see a detailed solution to problem 11.

            Click HERE to see a detailed solution to problem 12.

            Click HERE to see a detailed solution to problem 13.

            Click HERE to see a detailed solution to problem 14.

            Click HERE to see a detailed solution to problem 15.

            Click HERE or HERE to see a detailed solution to problem 16.

            Click HERE to see a detailed solution to problem 17.

            Click HERE to return to the original list of various types of calculus problems.

            Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address :

            A heartfelt "Thank you" goes to The MathJax Consortium and the online Desmos Grapher for making the construction of graphs and this webpage fun and easy.


            Trigonometric Substitution

            Consider the integral [int frac>.] At first glance, we might try the substitution $u=9-x^2$, but this will actually make the integral even more complicated!

            Let’s try a different approach:

            The radical $displaystylesqrt<9-x^2>$ represents the length of the base of a right triangle with height $x$ and hypotenuse of length $3$:

            For this triangle, $displaystylesin heta =frac<3>$, suggesting the substitution $x=3sin heta$. Then $displaystyle heta =arcsinleft(frac<3> ight)$, where we specify $-pi/2leq hetaleq pi/2$. Note that $dx=3cos heta, d heta$ and that $sqrt<9-x^2>=3cos heta$.

            With this change of variables, [int frac>= int frac <3cos heta , d heta><3cos heta>=int d heta = heta+C=arcsinleft(frac<3> ight)+C.]

            • The sketch of the triangle is very useful for determining what substitution should be made. Note, though, that the sketch only has meaning for $x > 0$ and $ heta > 0$.
            • It is important to be careful about how the angle $ heta$ is defined. With the restrictions on $ heta$ mentioned in the examples here, we avoid sign difficulties even when $x < 0$.

            There are two other trigonometric substitutions useful in integrals with different forms:

            Example

            Let’s evaluate [int frac<>>.] The radical $sqrt$ suggests a triangle with hypotenuse of length $x$ and base of length $2$:

            For this triangle, $displaystylesec heta =frac<2>$, we will try the substitution $x=2sec heta$. Then $displaystyle heta=sec^ <-1>left(frac<2> ight)$, where we specify leq heta<pi/2$ or $pileq heta <3pi/2$. Note that $dx=2sec heta an heta , d heta$ and that $sqrt=2 an heta$.

            We may also use a trigonometric substitution to evaluate a definite integral, as long as care is taken in working with the limits of integration:

            Example

            We will evaluate [int^1_<-1>frac<(1+x^2)^2>.] The factor $(1+x^2)$ suggests a triangle with base of length $1$ and height $x$:

            For this triangle, $ an heta =x$, so we will try the substitution $x= an heta$. Then $ heta = an^<-1>(x)$, where we specify $-pi/2< heta <pi/2$. Here, $dx=sec^2 heta , d heta$. Also, $sqrt<1+x^2>=sec heta$ so $(1+x^2)^2 = sec^4 heta$.

            There is often more than one way to solve a particular integral. A trigonometric substitution will not always be necessary, even when the types of factors seen above appear. With practice, you will gain insight into what kind of substitution will work best for a particular integral.

            Key Concepts

            Trigonometric substitutions are often useful for integrals containing factors of the form [(a^2-x^2)^n,qquadqquad (x^2+a^2)^n,qquad >qquad (x^2-a^2)^n.] The exact substitution used depends on the form of the integral:


            MathHelp.com

            The tangent is equal to 1 for on the first period. But this exercise wants the answer "in full generality". Obviously, I can't list out all of the solution values, because there are infinitely many of them. So I'll have to use a formula.

            From what I know about the graph of the tangent, I know that the tangent will equal 1 at 45° after every 180° . These solutions for are at 0° + 45°, 180° + 45°, 360° + 45° , and so forth. To give the answer "in full generality", I'll use a formula:

            Now I need to solve for x itself. I'll multiply through by 2 :

            In radians, the solution above would be etc and the general solution would be

            Solve 3tan 3 (x) &ndash 3tan 2 (x) &ndash tan(x) + 1 = 0 in full generality.

            The first equation solves, in the first period, as:

            The second solves, in the first period, as:

            To make the solution "general", I need to state the above solutions formulaically, to account for every period.

            The first solution is 45° more than a multiple of 180° , so (180n)° + 45° should do. The second solution is 30° more than a multiple of 180° and (because of the "plus / minus") also 30° less than that same multiple, so (180n)° ± 30° will cover this part.

            Solve on [0, 2&pi)

            When nothing looks like it's going to work, sometimes it helps to put everything in terms of sine and cosine. That process, applied to this equation, gives me:

            That's not a whole lot better. but the first two terms share a common factor of 2 . If I convert the last term to a common denominator with the third term, what will that give me?

            If I factor a 2 from the first two terms and the square root of 3 and a cosine from the second two terms, I'll get:

            Now I can take the common factor out front:

            Whew! That actually worked! Okay, now I need to solve the factors. The first factor solves as:

            This equation is true at x = 60° and, by the symmetry of the tangent curve, also at x = 180° + 60° = 240° . In radians, this is .

            The second factor solves as:

            Cosine takes on this value at x = 30° and, by the symmetry of the cosine curve, also at x = 360° &ndash 30° = 330° . In radians, this is . So my solution is:

            Solve ln(2 &ndash sin 2 (x)) = 0 on 0° < x < 360°

            The natural log (well, any log) is zero when the argument is 1 , so this gives me:

            From what I know of the sine wave, my solution is:

            Solve on [0, 2&pi)

            By nature of logarithms, the equivalent exponential equation is:

            The sine takes on this value at and also at . Then my solution is:

            Expect to need to factor (especially quadratics) in order to solve some trig equations, and also expect to need to use trig identities. Don't be afraid to try different methods sometimes your first impulse doesn't lead anywhere helpful, but your second guess might work fine. And pay particular attention to any oddly complex examples in your textbook, as these may hold hints about what tricks you will need, especially on the next test.

            You can use the Mathway widget below to practice solving trigonometric equations. Try the entered exercise, or type in your own exercise. (Unless you're told to solve "in full generality" remember to include an interval, as shown below.) Then click the button and, for best results, select "Solve over the Interval", to compare your answer to Mathway's.

            Note: The solver can only provide "exact" solutions, and sometimes any solution at all, if you're in radians. Use degrees at your own risk!

            (Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)


            4.2: The Trigonometric Series - Mathematics

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