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Question: Show that \[\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta \] ?...

Show that sin3θ=3sinθ4sin3θ\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta ?

Explanation

Solution

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 solution:
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
sin3θ=3sinθ4sin3θ\Rightarrow \,\,\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
sin3θ\Rightarrow \,\,\sin 3\theta ------(1)
The angle can be written in addition form i.e., 3θ=2θ+θ3\theta = 2\theta + \theta
Then equation (1) becomes
sin(2θ+θ)\Rightarrow \,\,\sin \left( {2\theta + \theta } \right)
As we know the sine sum formula sin(A+B)=sinAcosB+cosAsinBsin\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=2θA = \,2\theta and B=θB = \,\theta
Substitute A and B in formula then
sin2θcosθ+cos2θsinθ\Rightarrow \,\,\,sin2\theta \,cos\theta + cos2\theta \,sin\theta --------(2)
the double angle formula of cosine: cos(2θ)=12sin2θ\cos (2\theta ) = 1 - 2{\sin ^2}\theta and
the double angle formula of sine: sin(2θ)=2sinθcosθ\sin (2\theta ) = 2\sin \theta \cos \theta
substitute double angle formulas in equation (2), then
sin(2θ)cosθ+cos(2θ)sinθ\Rightarrow \,\,sin\left( {2\theta \,} \right)cos\theta + cos\left( {2\theta } \right)\,sin\theta
(2sinθcosθ)cosθ+(12sin2θ)sinθ\Rightarrow \,\,\,\left( {2\sin \theta \cos \theta } \right)cos\theta + \left( {1 - 2{{\sin }^2}\theta } \right)\,sin\theta
2sinθcos2θ+sinθ2sin3θ\Rightarrow \,\,\,2\sin \theta {\cos ^2}\theta + sin\theta - 2{\sin ^3}\theta \, ------(3)
Use the fundamental identity of trigonometry sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 cos2θ=1sin2θ\Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta .
Equation (3) becomes
2sinθ(1sin2θ)+sinθ2sin3θ\Rightarrow \,\,\,2\sin \theta \left( {1 - {{\sin }^2}\theta } \right) + sin\theta - 2{\sin ^3}\theta \,
On multiplying 2sinθ2\sin \theta
2sinθ2sin3θ+sinθ2sin3θ\Rightarrow \,\,\,2\sin \theta - 2{\sin ^3}\theta + sin\theta - 2{\sin ^3}\theta \,
On simplification, we get
3sinθ4sin3θ\Rightarrow \,\,\,3\sin \theta - 4{\sin ^3}\theta \,
= RHS
LHS=RHS\therefore \,\,LHS = RHS
Hence showed
sin3θ=3sinθ4sin3θ\sin 3\theta = 3\sin \theta - 4{\sin ^3}\theta
So, the correct answer is “ sin3θ=3sinθ4sin3θ\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.