Question

Prove it. this question is from trigonometry ​

Answer 1

Answer:

See below for step-by-step proof.

Step-by-step explanation:

$\textsf{Given expression}:$

$\cos^6 \theta+\sin^6 \theta$

$\textsf{Apply exponent rule} \quad a^{bc}=(a^b)^c:$

$\implies (\cos^2 \theta)^3+(\sin^2 \theta)^3$

$\textsf{Use the identity}\quad a^3+b^3=(a+b)^3-3ab(a+b):$

$\implies (\cos^2 \theta+\sin^2 \theta)^3-3 \cos^2 \theta\sin^2 \theta (\cos^2 \theta+\sin^2 \theta)$

$\textsf{Use the identity}\quad \sin^2 \theta + \cos^2 \theta=1:$

$\implies (1)^3-3 \cos^2 \theta\sin^2 \theta (1)$

$\implies 1-3 \cos^2 \theta\sin^2 \theta$

$\implies 1-3 (\cos \theta\sin \theta)^2$

$\textsf{Use the identity}\quad \sin 2 \theta=2 \sin \theta \cos \theta \implies \dfrac{1}{2}\sin 2 \theta= \sin \theta \cos \theta:$

$\implies 1-3 \left( \dfrac{1}{2}\sin 2 \theta\right)^2$

$\implies 1-\dfrac{3}{4} \sin^2 2 \theta$

$\textsf{Use the identity}\quad \sin^2 2\theta + \cos^2 2\theta=1 \implies \sin^2 2\theta=1-\cos^2 2\theta$

$\implies 1-\dfrac{3}{4} \left(1-\cos^2 2 \theta\right)$

$\implies 1-\dfrac{3}{4} +\dfrac{3}{4}\cos^2 2 \theta$

$\implies \dfrac{1}{4} +\dfrac{3}{4}\cos^2 2 \theta$

$\textsf{Factor out }\dfrac{1}{4}:$

$\implies \dfrac{1}{4}\left(1+3\cos^2 2 \theta\right)$

$\textsf{Use the identity}\quad \cos (4 \theta)=2 \cos^2 2\theta - 1 \implies \cos^2 2 \theta=\dfrac{1}{2}\cos 4 \theta+\dfrac{1}{2}:$

$\implies \dfrac{1}{4}\left(1+3\left(\dfrac{1}{2}\cos 4 \theta+\dfrac{1}{2}\right)\right)$

$\implies \dfrac{1}{4}\left(1+\dfrac{3}{2}\cos 4 \theta+\dfrac{3}{2}\right)$

$\implies \dfrac{1}{4}\left(\dfrac{5}{2}+\dfrac{3}{2}\cos 4 \theta\right)$

$\implies \dfrac{1}{4} \cdot \dfrac{1}{2}\left(5+3\cos 4 \theta\right)$

$\implies \dfrac{1}{8}\left(5+3\cos 4 \theta\right)$