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Question: Prove that \[\dfrac{\sin 16\theta }{\sin \theta }=16\cos \theta \cdot \cos 2\theta \cdot \cos 4\thet...

Prove that sin⁡16θsin⁡θ=16cos⁡θ⋅cos⁡2θ⋅cos⁡4θ.cos⁡8θ\dfrac{\sin 16\theta }{\sin \theta }=16\cos \theta \cdot \cos 2\theta \cdot \cos 4\theta .\cos 8\theta ?

Explanation

Solution

In the given question, we have been asked to prove the LHS of a given expression is equal to the RHS of the given expression. In order to solve the question, first we start by taking the LHS and simplify the expression in a way we can use the trigonometric identitysin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta . Then again we simplify the solved expression further in a way by using the identity. Solve and simplify the expression further until we get the expression that is equal to the RHS.

Complete step by step answer:
We have given that,
sin⁡16θsin⁡θ=16cos⁡θ⋅cos⁡2θ⋅cos⁡4θ.cos⁡8θ\dfrac{\sin 16\theta }{\sin \theta }=16\cos \theta \cdot \cos 2\theta \cdot \cos 4\theta .\cos 8\theta
Taking the LHS,
We have,
⇒sin⁡16θsin⁡θ\Rightarrow \dfrac{\sin 16\theta }{\sin \theta }
Rewrite the above expression as;
⇒sin⁡2(8θ)sin⁡θ\Rightarrow \dfrac{\sin 2\left( 8\theta \right)}{\sin \theta }
Using the trigonometric identity i.e. sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta
Here,
We have θ\theta = 8θ8\theta
Thus,
Applying the identity, we have
⇒2sin⁡8θcos⁡8θsin⁡θ\Rightarrow \dfrac{2\sin 8\theta \cos 8\theta }{\sin \theta }
Rewrite the above expression as;
⇒2sin⁡2(4θ)cos⁡8θsin⁡θ\Rightarrow \dfrac{2\sin 2\left( 4\theta \right)\cos 8\theta }{\sin \theta }
Using the trigonometric identity i.e. sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta
Here,
We have θ\theta = 4θ4\theta
Thus,
Applying the identity, we have
⇒2(2sin⁡4θcos⁡4θ)cos⁡8θsin⁡θ\Rightarrow \dfrac{2\left( 2\sin 4\theta \cos 4\theta \right)\cos 8\theta }{\sin \theta }
Rewrite the above expression as;
⇒4(sin⁡2(2θ))⋅cos⁡4θ⋅cos⁡8θsin⁡θ\Rightarrow \dfrac{4\left( \sin 2\left( 2\theta \right) \right)\cdot \cos 4\theta \cdot \cos 8\theta }{\sin \theta }
Using the trigonometric identity i.e. sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta
Here,
We have θ\theta = 2θ2\theta
Thus,
Applying the identity, we have
⇒4(2sin⁡2θcos⁡2θ)⋅cos⁡4θ⋅cos⁡8θsin⁡θ\Rightarrow \dfrac{4\left( 2\sin 2\theta \cos 2\theta \right)\cdot \cos 4\theta \cdot \cos 8\theta }{\sin \theta }
Rewrite the above expression as;
⇒8(sin⁡2θ)cos⁡2θ⋅cos⁡4θ⋅cos⁡8θsin⁡θ\Rightarrow \dfrac{8\left( \sin 2\theta \right)\cos 2\theta \cdot \cos 4\theta \cdot \cos 8\theta }{\sin \theta }
Using the trigonometric identity i.e. sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta
Thus,
Applying the identity, we have
⇒8(2sin⁡θcos⁡θ)cos⁡2θ⋅cos⁡4θ⋅cos⁡8θsin⁡θ\Rightarrow \dfrac{8\left( 2\sin \theta \cos \theta \right)\cos 2\theta \cdot \cos 4\theta \cdot \cos 8\theta }{\sin \theta }
Simplifying the above, we get
⇒16sin⁡θ⋅cos⁡θcos⁡2θ⋅cos⁡4θ⋅cos⁡8θsin⁡θ\Rightarrow \dfrac{16\sin \theta \cdot \cos \theta \cos 2\theta \cdot \cos 4\theta \cdot \cos 8\theta }{\sin \theta }
Cancelling out the common terms, we get
⇒16cos⁡θ⋅cos⁡2θ⋅cos⁡4θ⋅cos⁡8θ=RHS\Rightarrow 16\cos \theta \cdot \cos 2\theta \cdot \cos 4\theta \cdot \cos 8\theta =RHS
Therefore,
⇒sin⁡16θsin⁡θ=16cos⁡θ⋅cos⁡2θ⋅cos⁡4θ.cos⁡8θ\Rightarrow \dfrac{\sin 16\theta }{\sin \theta }=16\cos \theta \cdot \cos 2\theta \cdot \cos 4\theta .\cos 8\theta
Hence proved.

Note: In order to solve these types of questions, you should always need to remember the properties of trigonometric and the trigonometric identities as well. It will make questions easier to solve. It is preferred that while solving these types of questions we should carefully examine the pattern of the given function and then you would apply the formulas according to the pattern observed. As if you directly apply the formula it will create confusion ahead and we will get the wrong answer.