Trigonometric Identities (problem solving)

1) If cos 2A = $tan^2$ B then find the value of $tan^2$ A in terms of B.

Sol:

Given, cos 2A = $tan^2$ B

Use multiple angle formula of cos, i.e. cos 2A = $\frac{1- tan^2\ A}{1+ tan^2\ A}$

$\frac{1- tan^2\ A}{1+ tan^2\ A}$ = $tan^2$ B

1- $tan^2$ A = $tan^2$ B (1+ $tan^2$ A)

1- $tan^2$ A = $tan^2$ B + $tan^2$ B $tan^2$ A

1- $tan^2$ B = $tan^2$ A + $tan^2$ B $tan^2$ A

1- $tan^2$ B = $tan^2$ A(1- $tan^2$ B)

$\frac{1- tan^2\ B}{1+ tan^2\ B}$ = $tan^2$ A

Again use the formula cos 2A = $\frac{1- tan^2\ A}{1+ tan^2\ A}$

cos 2B = $tan^2$ A

Therefore $tan^2$ A = cos 2B

2) If a cos A - b sin A = c then find the value of a sin A + b cos A

Sol:

Given, a cos A - b sin A = c

Squaring on both sides

$(a cos A - b sin A)^2$ = $c^2$

$a^2\ cos^2$ A + $b^2\ sin^2$ A -2ab sin A cos A = $c^2$

$a^2\ (1-sin^2$ A) + $b^2\ (1-sin^2$ A) -2ab sin A cos A = $c^2$

$a^2$ - $a^2\ sin^2$ A + $b^2$ - $b^2\ sin^2$ A -2ab sin A cos A = $c^2$

$a^2$ + $b^2$ - ( $a^2\ sin^2$ A + $b^2\ sin^2$ A + 2ab sin A cos A) = $c^2$

(a sin A + b cos A $)^2$ = $a^2$ + $b^2$ - $c^2$

a sin A + b cos A = $\pm \sqrt{a^2 + b^2 - c^2}$

3) If sin A + cosec A = 2 then find the value of sin $^20$ A + cosec $^20$ A

Sol:

sin A + cosec A = 2

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

Sin $^2$ A + 1 = 2 sin A

Sin $^2$ A + 1 - 2 sin A = 0

(sin A - 1 $)^2$ = 0

Sin A = 1

So cosec A = 1

sin $^20$ A + cosec $^20$ A =1+1

sin $^20$ A + cosec $^20$ A = 2

4) If sin A + sin $^2$ A + sin $^3$ A = 1 find the value of cos $^6$ A - 4cos $^4$ A + 8cos $^2$

Sol:

Given, sin A + sin $^2$ A + sin $^3$ A = 1

sin A + sin $^3$ A = 1- sin $^2$ A

sin A + sin $^3$ A = cos $^2$ A

Square on both sides

( sin A + sin $^3$ A $)^2$ = cos $^4$ A

sin $^2$ A + 2sin $^4$ A + sin $^6$ A = cos $^4$ A

1-cos $^2$ A + 2(1-cos $^2$ A $)^2$ + (1-cos $^2$ A $)^3$ = cos $^4$ A

1-cos$^2$ A + 2-4cos $^2$ A + 2cos $^4$ A + 1- 3cos $^2$ A + 3cos $^4$ A - cos $^6$ A = cos $^4$ A

4 - cos $^6$ A + 4cos $^4$ A - 8cos $^2$ = 0

Therefore cos $^6$ A - 4cos $^4$ A + 8cos $^2$= 0

5) if sin A + cos A = m and sin $^3$ A + cos $^3$ A = n, then prove that m$^3$ -3m+2n=0

sol:

Given,

sin A + cos A = m and sin $^3$ A + cos $^3$ A = n

Consider the second equation

sin $^3$ A + cos $^3$ A = n

Use the algebraic formula a $^3$ A + b $^3$ A = (a+b)( a $^2$ - ab + b $^2$ )

(sin A + cos A)( sin $^2$ A - sin A cos A + cos $^2$ A) = n

But it is given that sin A + cos A = m and we know that sin $^2$ A + cos $^2$ A = 1

m(1 - sin A cos A) = n ————–(1)

Now let’s consider the given equation sin A + cos A = m and lets square on both sides of the equation

(sin A + cos A $)^2$ = m $^2$

sin $^2$ A + 2sin A cos A + cos $^2$ A = m $^2$

we know that sin $^2$ A + cos $^2$ A = 1

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

Now from (1) and (2)

m(1 - $\frac{m^2 -1}{2}$) = n

2m - m $^3$ +m = 2n

m$^3$ -3m+2n=0