Prove the trigonometric identity: $\frac{\sin heta - 2 \sin^3 heta}{2 \cos^3 heta - \cos heta} = an heta$. S

Question

Prove the trigonometric identity: $\frac{\sin 	heta - 2 \sin^3 	heta}{2 \cos^3 	heta - \cos 	heta} = 	an 	heta$. Show all steps clearly.

Accepted Answer

To prove: $\frac{\sin 	heta - 2 \sin^3 	heta}{2 \cos^3 	heta - \cos 	heta} = 	an 	heta$ 
Take the Left Hand Side (LHS): $	ext{LHS} = \frac{\sin 	heta - 2 \sin^3 	heta}{2 \cos^3 	heta - \cos 	heta}$ 
Factor out $\sin 	heta$ from the numerator and $\cos 	heta$ from the denominator: $	ext{LHS} = \frac{\sin 	heta (1 - 2 \sin^2 	heta)}{\cos 	heta (2 \cos^2 	heta - 1)}$ 
Using the fundamental trigonometric identity $\sin^2 	heta + \cos^2 	heta = 1$, we can substitute $1 = \sin^2 	heta + \cos^2 	heta$ in the brackets, or use $\cos^2 	heta = 1 - \sin^2 	heta$:
Let us substitute $\cos^2 	heta = 1 - \sin^2 	heta$ in the denominator:
Denominator term $= 2(1 - \sin^2 	heta) - 1$ $= 2 - 2 \sin^2 	heta - 1$ $= 1 - 2 \sin^2 	heta$ 
Now, substitute this back into the LHS expression: $	ext{LHS} = \frac{\sin 	heta (1 - 2 \sin^2 	heta)}{\cos 	heta (1 - 2 \sin^2 	heta)}$ 
Cancel out the common term $(1 - 2 \sin^2 	heta)$ from both numerator and denominator: $	ext{LHS} = \frac{\sin 	heta}{\cos 	heta}$ 
We know that $\frac{\sin 	heta}{\cos 	heta} = 	an 	heta$, which is equal to the Right Hand Side (RHS). $	ext{LHS} = 	ext{RHS}$
Hence proved.

Introduction to Trigonometry

❓ Sibling Questions in this Chapter