Question

plz prove this ques its an urgent​

Answer 1

Step-by-step explanation:

Hope this attachment will help you

Answer 2

Question ;

to proof

$\cos(4x) = 1 - 8 \sin {}^{2} (x) + 8 \sin {}^{4} (x)$

solution ;

we know that

$\cos(2x) = 2 \cos {}^{2} (x) - 1$

replace 2x by

$\cos2(2x) = 2 \cos {}^{2} (2x) - 1$

then ,

$\cos(4x) = 2 \cos {}^{2} (2x) - 1$

by using cos2x formula

$\cos(4x) = 2 (\cos(2x) - 1) {}^{2} - 1$

using formula ;

(a-b)² = a²-b²

$= 2((2 \cos(x) ) {}^{2} + 1 {}^{2} - 2(2 \cos {}^{2} (x) \times 1) - 1$

$= 2(4 \cos {}^{4} (x) + 1 - 4 \cos {}^{2} (x) ) - 1$

$= 8 \cos {}^{4} (x) + 2 - 8 \cos {}^{2} (x) - 1$

$= 8 \cos {}^{4} (x) - 8 \cos {}^{2} (x) + 1$

$= 8 \cos {}^{2} (x) ( \cos {}^{2} (x) - 1) + 1$

$= - 8 \cos {}^{2} (x) (1 - \cos {}^{2} x ) + 1$

$= 1 - 8 \cos {}^{2} (x) \sin {}^{2} (x)$

we know that

sin²x+cos²x= 1

then ,

$= 1 - 8(1 - \sin {}^{2} (x) ) \sin {}^{2} (x)$

therefore ;

$\cos(4x) = 1 - 8 \sin{}^{2} + 8 \sin {}^{4} (x)$

I hope it helps you!