Question

Prove it!

.........!​

Answer 1

Answer:

$\cos {}^{3} (x) \cos(3x) + \sin {}^{3} (x) \sin(3x) \\ \\ =\cos {}^{3} (x) \{ 4\cos {}^{3} (x) - 3 \cos(x) \} \\ + \sin {}^{3} (x) (3\sin(x) - 4 \sin {}^{3} (x) ) \\ \\ = 4 \cos {}^{6} (x) + 3 { \sin {}^{4} (x) }\\ - 3 \cos {}^{4} (x) - 4 \sin {}^{6} (x) \\ \\ let \: \: y = \cos {}^{2} (x) \\ \implies \: \sin {}^{2} (x) =1 - \cos {}^{2} (x) = 1 - y \\ \\ put \: values \: in \: \: equation \\ \\ = 4 {y}^{3} + 3(1 - y) {}^{2} \\- 3 {y}^{2} - 4(1 - y) {}^{3} \\ \\ = 4( {y}^{3} - (1 - y) {}^{3} ) \\+ 3( {(1 - y)}^{2} - (y) {}^{2} ) \\ \\ do \: \: full \: \: expansion \: \: or \: \: use \: \\ basic \: \: identities \: \\ \\ = 4(y - 1 + y)( {y}^{2} + {(1 - y)}^{2} + y(1 - y)) + 3( 1 - y - y)(1 - y+ y) \\ \\ = 4(2y - 1)( {y}^{2} + {y}^{2} + 1 - 2y + y - {y}^{2} ) + 3(1 - 2y) \\ \\ = (2y - 1) \{4({y}^{2} - y + 1 \} - 3) \\ \\ = (2y - 1)(4 {y}^{2} - 4y + 1) \\ \\ = (2y - 1)(2y - 1) {}^{2} \\ \\ = (2y - 1) {}^{3} \\ \\ since \: \: y = \cos {}^{2} (x) \\ \\ = (2 \cos {}^{2} (x) - 1) {}^{2} \\ \\ = \cos {}^{3} (2x) \\ \\ = rhs$

Answer 2

Answer:

not interested

Step-by-step explanation:

The answer is here,

Given that,

= > \: \frac{a + ib}{c + id} = p + iq

We can replace the " i " as " -i ".

= > \frac{a - ib}{c - id} = p - iq