Answer to Question #137552 in Complex Analysis for Axwell Alesso Lee

Question

Express cos^4
θ sin^3
θ in terms of multiples of angles.

Answer 1

According to Euler's formula:

$z^{n}+z^{-n} = 2cosnx$

$16 \cdot cos^{4}x = (2 \cdot cosx)^{4} = (z + z^{-1})^{4}$

Thus, applying binomial formula:

$128 \cdot cos^{4}x = z^{4} + 4\cdot z^{2} + 6 + 4\cdot z^{-2} + z^{-4} = 2cos4x + 8 cos2x + 6$

Thus, $cos^{4}x = \frac{2\cdot cos4x + 8\cdot cos2x + 6}{128}= \frac{cos4x + 4\cdot cos2x + 3}{64}$

According to Euler's formula:

$z^{n}-z^{-n} = 2i\cdot sinnx$

$8 \cdot sin^{3}x = i(2i \cdot sinx)^{3} = i(z - z^{-1})^{3}$

Thus, applying binomial formula:

$8 \cdot sin^{3}x = i \cdot (z^{3} - 3\cdot z + 3\cdot z^{-1} - z^{-3}) = i \cdot (2i \cdot sin3x -6i \cdot sinx) = 6 sinx - 2sin3x$

Thus, $sin^{3}x = \frac{3sinx - sin3x}{4}$

$sinx \cdot cos4x = \cfrac{1}{2} \cdot (sin5x - sin 3x)$

$sinx \cdot cos2x = \cfrac{1}{2} \cdot (sin3x - sin x)$

$sin3x \cdot cos4x = \cfrac{1}{2} \cdot (sin7x - sin x)$

$sin3x \cdot cos2x = \cfrac{1}{2} \cdot (sin5x + sin x)$

Thus, $cos^{4}x \cdot sin^{3}x = \cfrac{1}{256}(3sinx - sin3x) \cdot(cos4x + 4 \cdot cos2x +3) =$

$=\cfrac{1}{512}(3sin5x - 3sin3x - sin7x + sinx +12 sin 3x - 12sinx -4sin5x - 4sinx + 9sinx - 3sin3x)=\cfrac{1}{512}\cdot (6sin3x - sin5x - sin7x - 6sinx)$