Question

How do you prove \[\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta \] ?

Answer 1

Hint:The question is related to trigonometry which is a sine of triple angle, here we have to prove the given identity by the standard trigonometric formula sine sum and the double angle formula of sine and cosine and on further simplification we get the required proof of the given identity.

Complete step by step answer:
Trigonometric ratios: Some ratios of the sides of a right-angle triangle with respect to its acute angle called trigonometric ratios of the angle. Let us consider the given trigonometry identity
\[ \sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta \]
Simplify the Left Hand Side (LHS) part until the RHS part becomes the same. Then LHS will be equal to RHS which is written as: LHS= RHS it means the identity was proved.Consider the LHS,
\[\sin 3\theta \]------(1)
The angle can be written in addition form i.e., \[3\theta = 2\theta + \theta \]
Then equation (1) becomes
\[\sin \left( {2\theta + \theta } \right)\]
As we know the sine sum formula \[sin\left( {A + B} \right) = sinA\,cosB + cosA\,sinB\]. This formula is also known as trigonometry function for sum of two angles. Where A and B represents the angles.
Here \[A = \,2\theta \] and \[B = \,\theta \]

Substitute A and B in formula then
\[\sin2\theta \,cos\theta + cos2\theta \,sin\theta \]--------(2)
The double angle formula of cosine: \[\cos (2\theta ) = 1 - 2{\sin ^2}\theta \] and the double angle formula of sine: \[\sin (2\theta ) = 2\sin \theta \cos \theta \]
substitute double angle formulas in equation (2), then
\[ \Rightarrow \,\,sin\left( {2\theta \,} \right)cos\theta + cos\left( {2\theta } \right)\,sin\theta \]
\[ \Rightarrow \,\,\,\left( {2\sin \theta \cos \theta } \right)cos\theta + \left( {1 - 2{{\sin }^2}\theta } \right)\,sin\theta \]
\[ \Rightarrow \,\,\,2\sin \theta {\cos ^2}\theta + sin\theta - 2{\sin ^3}\theta \,\]------(3)

Use the fundamental identity of trigonometry \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]\[ \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta \].
Equation (3) becomes
\[2\sin \theta \left( {1 - {{\sin }^2}\theta } \right) + sin\theta - 2{\sin ^3}\theta \,\]
On multiplying \[2\sin \theta \]
\[2\sin \theta - 2{\sin ^3}\theta + sin\theta - 2{\sin ^3}\theta \,\]
On simplification, we get
\[3\sin \theta - 4{\sin ^3}\theta \,\]
\[\therefore \,\,LHS = RHS\]

Hence proved \[\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta \].

Note:The question is involving the trigonometry terms. Here we must know about the trigonometry functions of sum of two angles and double angle trigonometry ratios. By using the above formulas, we are going to simplify the given trigonometric function. While simplifying we should take care of signs.