Question

IIT JEE question

find the value of
cos³(π/8) cos( 3π/8)
+ sin ³(π/8) sin ( 3π/8)



​

Answer 1

$\huge{\blue{\underline{\purple{\mathbb{Trigonometry}}}}}$

forumlas used ;

cos (π/2- X) = cos X

Sin (π/2- X) = sinx

sinx²+ cos x²= 1

and 2 sin A sin B = sin (A+B) + sin ( A-B)

❇️step by step explanation

cos³(π/8) cos( 3π/8)

+ sin ³(π/8) sin ( 3π/8)

= cos ³(π/8) cos (π/2 - π/8) +

sin³(π /8) sin (π/2- π/8)

= cos³(π/8) sin ( π/8)

+ sin ³(π/8) cos ( π/8)

= sin (π /8) cos (π /8) [ sin²(π/8) + cos²( π/8) ]

{ since sin x² + cos x²= 1 }

then ,

= sin (π/8) cos (π/8)

multiply and divide by 2

=

$\frac{1}{2} (2 \sin( \frac{\pi}{8} ) \cos( \frac{\pi}{8} )$

now use formula

[ 2 sin A sin B = sin (A+B) + sin ( A-B)]

then , we get

=

$= \frac{1}{2} ( \sin( \frac{\pi}{4} ) \times \sin(0) )$

$= \frac{1}{2 \sqrt{2} }$

• so the value of

cos³(π/8) cos( 3π/8)

+ sin ³(π/8) sin ( 3π/8) = 1/(2√ 2)

I hope it helps you

❇️follow me

Answer 2

cos³(π/8) cos( 3π/8)

+ sin ³(π/8) sin ( 3π/8)

= cos ³(π/8) cos (π/2 - π/8) +

sin³(π /8) sin (π/2- π/8)

= cos³(π/8) sin ( π/8)

+ sin ³(π/8) cos ( π/8)

= sin (π /8) cos (π /8) [ sin²(π/8) + cos²( π/8) ]

{ since sin x² + cos x²= 1 }

then ,

= sin (π/8) cos (π/8)

multiply and divide by 2

=

$\frac{1}{2} (2 \sin( \frac{\pi}{8} ) \cos( \frac{\pi}{8} )21(2sin(8π)cos(8π)$

now use formula

[ 2 sin A sin B = sin (A+B) + sin ( A-B)]

then , we get

=

= $\frac{1}{2} ( \sin( \frac{\pi}{4} ) \times \sin(0) )=21(sin(4π)×sin(0))$

= $\frac{1}{2 \sqrt{2} }=221$