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Powers of Trig Functions

A very important subject of mathematics, dealing with relationships between ratios of sides and angles of right-angled triangle, is known as trigonometry. There are total six trigonometric functions studied by us. These are listed as sine, cosecant, cosine, secant, tangent, cotangent and are written in short as sin, cosec, cos, sec, tan, cot respectively. The sine and cosecant functions are reciprocal to each other. Similarly, cosine and sec are reciprocal; while tan and cot are reciprocal to each other.

In mathematics, we often deal with powers of trigonometric functions. The trig functions are represented in power or exponent form as:
(sin x)$^{n}$ = sin$^{n}$x

(cos x)$^{n}$ = cos$^{n}$x

(tan x)$^{n}$ = tan$^{n}$x

(cosec x)$^{n}$ = cosec$^{n}$x

(sec x)$^{n}$x = sec$^{n}$x

(cot x)$^{n}$x = cot$^{n}$x

Where, n is a number.

There are various formulae defined in trigonometry related to powers of trigonometric functions. There are many differentiation and integration problems that deal with powers of trig functions. Let us study about these formulae and some sample problems based on powers of trigonometric functions.

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Power Formulae

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There are various formulae related to powers of trigonometric functions. These are listed below:

Pythagorean Trigonometric Identities:

i)
$sin^{2} A + cos^{2} A = 1$; which given rise two more results by manipulating trig functions:

a) $sin^{2} A = 1 - cos^{2} A$

b) $cos^{2} A =  1 - sin^{2} A$

ii) $1 + tan^{2} A = sec^{2} A$ leads to two more formulae:


a) $tan^{2} A = sec^{2} A - 1$

b) $sec^{2} A - tan^{2} A = 1$

iii) $1 + cot^{2} A = cosec^{2} A$ which also leads to two more result:

a) $cot^{2} A = cosec^{2} A - 1$

b) $cosec^{2} A - cot^{2} A = 1$

Other Formulae:

i) $sin^{2} A$ = $\frac{1}{2}$ $(1-cos 2A)$

ii) $sin^{3} A$ = $\frac{1}{4}$ $(3 sin A - sin 3A)$

iii) $sin^{4} A$ = $\frac{1}{8}$ $(3 - 4 cos 2A - cos 4A)$

iv) $cos^{2} A$ = $\frac{1}{2}$ $(1+cos 2A)$

v) $cos^{3} A$ = $\frac{1}{4}$ $(3 cos A + cos 3A)$

vi) $sin^{4} A$ = $\frac{1}{8}$ $(3 + 4 cos 2A + cos 4A)$

vii) $tan^{2} A$ = $\frac{1-cos 2A}{1+ cos 2A}$

Integration Formulae

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There are few formula which are used while dealing with integration of high powers of trig functions:

Lets have a look at the formula for integration of n$^{th}$ power of sine.

$\int sin^{n}x\ dx$

= $\int sin^{n-1}x\ sin x\ dx$

= $\int sin^{n-1}x\ (-cos\ x) - \int [(-cos\ x) (n-1)sin^{n-2}x \ cos\ x] dx$

= $- \int sin^{n-1}x\ cos\ x + (n-1) \int [cos^{2}\ x\ sin^{n-2}x] dx$

= $- \int sin^{n-1}x\ cos\ x + (n-1) \int [(1-sin^{2}\ x)\ sin^{n-2}x] dx$

$\int sin^{n}x\ dx = - \int sin^{n-1}x\ cos\ x + (n-1) \int [sin^{n-2}x-sin^{n}\ x] dx$

Combining and finding the value of $\int sin^{n}x\ dx$, we get the following result:
$\int sin^{n}x\ dx$ = -$\frac{1}{n}$ $(sin^{n-1}x\ cos\ x)$ + $\frac{n-1}{n}$ $\int sin^{n-2}x\ dx$Similarly, there are other formulae too, all of which are known as reduction formulae. These are as follows:
$\int cos^{n}x\ dx$ = $\frac{1}{n}$ $(cos^{n-1}x\ sin\ x)$ + $\frac{n-1}{n}$ $\int cos^{n-2}x\ dx$

$\int tan^{n}x\ dx$ = $\frac{1}{n-1}$ $(tan^{n-1}x) - \int tan^{n-2}x\ dx$

$\int sec^{n}x\ dx$ = $\frac{1}{n-1}$ $(sec^{n-2}x\ tan\ x)$ + $\frac{n-2}{n-1}$ $\int sec^{n-2}x\ dx$

Problems

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Few problems related to powers of trig functions are given below:

Problem 1: Differentiate sec$^{3}$A.

Solution: $\frac{d}{dx}$ $sec^{3} A$

= 3 $sec^{2}A$ $\frac{d}{dx}$ $sec A$

= 3 $sec^{2}A sec A\ tan A$

= 3 $sec^{3}A \ tan A$

Problem 2: Integrate $cos^{2} A$ with respect to A. Verify the answer with the use of reduction formula.

Solution : $\int cos^{2} A\ dA$

= $\int$ $\frac{1}{2}$ $(1 + cos 2A)\ dA$

=$\frac{1}{2}$ ( $\int dA + \int cos 2A\ dA$)

= $\frac{1}{2}$ A + $\frac{1}{4}$ $sin 2A + c$

= $\frac{A}{2}$ + $\frac{2 sinA cos A}{4}$ + $c$

The reduction formula, $\int cos^{n}x\ dx$ = $\frac{1}{n}$ $(cos^{n-1}x\ sin\ x)$ + $\frac{n-1}{n}$ $\int cos^{n-2}x\ dx$
Plug n = 2 and x = A

$\int cos^{2}A\ dA$ = $\frac{1}{2}$ (cos^{2-1}A\ sin\ A)$ + $\frac{2-1}{2}$ $\int cos^{2-2}A\ dA$

= $\frac{1}{2}$ $(cos\ A\ sin\ A)$ + $\frac{1}{2}$ $\int  dA$

= $\frac{1}{2}$ $(cos\ A\ sin\ A) +$\frac{1}{2}$ $A + c$

which is same as above answer.

Problem 3: Find the integral of $sin^{3}x\  cos^{5}x$ with respect to x.

Solution: $\int sin^{3}x\ cos^{5}x\ dx$

= $\int sin^{3}x\ cos^{4}x\ cos x\ dx$

= $\int sin^{3}x\ (1-sin^{2}x)^{2}\ cos x\ dx$

= $\int sin^{3}x\ (1 + sin^{4}x - 2 sin^{2} x)\ cos x\ dx$

= $\int (sin^{3}x\ + sin^{7}x - 2 sin^{5} x)\ cos x\ dx$

Substitute sin x = t

then cos x dx = dt

= $\int (t^{3}\ + t^{7} - 2 t^{5})\ dt$

= $\frac{t^{4}}{4}$ + $\frac{t^{8}}{8}$ - 2 $\frac{t^{6}}{6}$ + $c$

= $\frac{sin^{4}x}{4}$  + $\frac{sin^{8}x}{8}$ - $\frac{sin^{6}x}{3}$ + $c$

Problem 4: Find $\int sin^{3}x\ dx$ using reduction formula.

Solution: $\int sin^{3}x\ dx$

The suitable reduction formula is:

$\int sin^{n}x\ dx$ = -$\frac{1}{n}$ $(sin^{n-1}x\ cos\ x)$ + $\frac{n-1}{n}$ $\int sin^{n-2}x\ dx$

Substituting n = 3 in it, we get

$\int sin^{3}x\ dx$ = -$\frac{1}{3}$ $(sin^{3-1}x\ cos\ x)$ + $\frac{3-1}{3}$ $\int sin^{3-2}x\ dx$

= -$\frac{1}{3}$ $(sin^{2}x\ cos\ x)$ + $\frac{2}{3}$ $\int sin\ x\ dx$

= -$\frac{1}{3}$ $(sin^{2}x\ cos\ x)$ - $\frac{2}{3}$ $cos\ x + c$
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