# DIFFERENTIATION BY USING TRIGONOMETRIC SUBSTITUTION

## Trigonometric Identities

sin 2A  =  2 Sin A cos A

cos 2A  =  cos² A - Sin² A

tan 2A  =  2tan A/(1-tan² A)

cos 2A  =  1 - 2Sin² A

cos 2A  =  2Cos² A - 1

sin 2A  =  2 tan A/(1+tan² A)

cos 2A  =  (1-tan² A)/(1+tan² A)

sin²A  =  (1-cos2A)/2

cos²A  =  (1+cos2A)/2

tan²A  =  (1-Cos2A)/(1+Cos2A)

sin 3A  =  3 sin A - 4 sin³A

cos 3A  =  4 cos³A - 3 cos A

tan 3A  =  (3 tan A - tan³ A)/(1-3tan²A)

sin A  =  2 Sin (A/2) cos (A/2)

cos A  =  cos² (A/2) - Sin² (A/2)

tan A  =  2 tan (A/2)/[1-tan² (A/2)]

cos A  =  1 - 2sin² (A/2)

cos A  =  2cos² (A/2) - 1

sin A  =  2 tan (A/2)/[1 + tan² (A/2)]

cos A  =  [1 - tan²(A/2)]/[1 + tan² (A/2)]

sin²A/2  =  (1 - cos A)/2

cos²A/2  =  (1 + cos A)/2

tan²(A/2)  =  (1 - cos A)/(1 + cos A)

Let us look at some examples to understand how to differentiate a function using trigonometric substitution.

Example 1 :

Differentiate the following

tan-1[√(1 - cos x)/(1+cosx)]

Solution :

Let y  =  tan-1[√(1 - cos x)/(1+cosx)]

Trigonometric formula for (1-cos x)/(1+cosx)  =  tan2(x/2)

By applying trigonometric formula, we may find derivatives easily.

y  =  tan-1[√tan2(x/2)]

y = x/2

Differentiating with respect to "x"

dy/dx  =  1/2

Example 2 :

Differentiate the following

tan-1[6x/1-9x2]

Solution :

Let y  =  tan-1[6x/1-9x2]

y  =  tan-1[2(3x/1-(3x)2]

Let 3x  =  tan θ

y  =  tan-1[2 tan θ /1-tan2θ]

tan θ /1-tan2θ  =  tan 2θ

y  =  tan-1[tan 2θ]

y = 2θ

dy/dx  =  2 (dθ/dx)  ------(1)

3x  =  tan θ

3(1)  =  sec2θ(dθ/dx)

3/sec2θ  =  (dθ/dx)

Applying dθ/dx  =  3/sec2θ in (1)

dy/dx  =  2 (3/sec2θ

=  2(3)/(1  + tan2θ)

=  6/(1 + (3x)2)

=  6/(1 + 9x2)

Example 3 :

Differentiate the following

cos (2tan-1[√(1-x)/(1+x)])

Solution :

Let y  =  cos (2tan-1[√(1-x)/(1+x)])

let x = cos θ

Differentiating with respect to x, we get

dx/dθ  =  -sin θ

y  =  cos (2tan-1[√(1-cosθ)/(1+cosθ)])

y  =  cos (2tan-1[√tan2(θ/2])

y  =  cos (2tan-1[tan(θ/2])

y  =  cos θ

Differentiating with respect to θ.

dy/dx  =  -sin θ (dθ/dx)    -----(1)

Now let us apply the value of dx/dθ  =  -sin θ in (1)

dy/dx  =  -sin θ (-1/sinθ)

dy/dx  =  1

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