Answer to Question #176570 in Trigonometry for Yusuf Idris

Question

Show that cos³θ +sin³θ =1/4(cos³θ + 3cosθ - sin³θ + 3sinθ)

Answer 1

• It is a rearrangement of several trigonometric identities.
• As it is known,

$\qquad\qquad \begin{aligned} \small \sin3\theta &=\small 3\sin\theta-4\sin\theta^3\\ \small \cos3\theta&=\small 4\cos^3\theta -3\cos\theta \end{aligned}$

• Then by rearranging,

$\qquad\qquad \begin{aligned} \small 4\cos^3\theta &=\small \cos3\theta+3\cos\theta\\ \small \cos^3\theta&=\small \frac{1}{4}\cdot(\cos3\theta+3\cos\theta)\cdots\cdots(1)\\ \\ \small 4\sin^3\theta&=\small 3\sin\theta-\sin^3\theta\\ \small \sin^3\theta&=\small \frac{1}{4}\cdot(3\sin\theta-\sin^3\theta)\cdots\cdots(2)\\ \\ &\small\text{Then by (1)+(2)}\\ \\ \small \sin^3\theta+\cos^3\theta &=\small \frac{1}{4}(3\sin\theta-\sin^3\theta) +\frac{1} {4}(\cos3\theta+3\cos\theta)\\ &=\small \frac{1}{4}\Big[3\cos\theta+\cos3\theta+3\sin\theta-\sin^3\theta\Big] \end{aligned}$