Table of Contents
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
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