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Question: \(4\sin \alpha .\sin (60 - \alpha ).\sin (60 + \alpha ) = \sin 3\alpha \)...

4sin⁡α.sin⁡(60−α).sin⁡(60+α)=sin⁡3α4\sin \alpha .\sin (60 - \alpha ).\sin (60 + \alpha ) = \sin 3\alpha

Explanation

Solution

To prove the given trigonometric equation, solve the L.H.S and make it equal to R.H.S. to solve this we can apply the trigonometric formula given as:
⇒sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\Rightarrow \sin (A \pm B) = \sin A\cos B \pm \cos A\sin B …….(1)
Here in this question A=60∘A = {60^ \circ }and B=αB = \alpha . Put this in formula and solve the L.H.S

Complete step by step solution:
We are given with a trigonometric equation and we have to prove this equation true i.e.
⇒4sin⁡α.sin⁡(60∘−α).sin⁡(60∘+α)=sin⁡3α\Rightarrow 4\sin \alpha .\sin ({60^ \circ } - \alpha ).\sin ({60^ \circ } + \alpha ) = \sin 3\alpha
Taking the left-hand side and take last two terms of L.H.S in bracket we get,
⇒4sin⁡α.sin⁡(60∘−α).sin⁡(60∘+α)=4sin⁡α(sin⁡(60∘−α).sin⁡(60∘+α))\Rightarrow 4\sin \alpha .\sin ({60^ \circ } - \alpha ).\sin ({60^ \circ } + \alpha ) = 4\sin \alpha (\sin ({60^ \circ } - \alpha ).\sin ({60^ \circ } + \alpha )) ……….(2)
By applying the trigonometric formula given in (1) on sin⁡(60∘−α)\sin ({60^ \circ } - \alpha ) and sin⁡(60∘+α)\sin ({60^ \circ } + \alpha ) we get,
⇒sin⁡(60∘−α)=sin⁡60∘cos⁡α−cos⁡60∘sin⁡α\Rightarrow \sin ({60^ \circ } - \alpha ) = \sin {60^ \circ }\cos \alpha - \cos {60^ \circ }\sin \alpha
⇒sin⁡(60∘+α)=sin⁡60∘cos⁡α+cos⁡60∘sin⁡α\Rightarrow \sin ({60^ \circ } + \alpha ) = \sin {60^ \circ }\cos \alpha + \cos {60^ \circ }\sin \alpha
By putting these values in the equation 2 we get,
⇒4sin⁡α(sin⁡(60∘−α).sin⁡(60∘+α))=4sin⁡α((sin⁡60∘cos⁡α−cos⁡60∘sin⁡α)(sin⁡60∘cos⁡α+cos⁡60∘sin⁡α))\Rightarrow 4\sin \alpha (\sin ({60^ \circ } - \alpha ).\sin ({60^ \circ } + \alpha )) = 4\sin \alpha ((\sin {60^ \circ }\cos \alpha - \cos {60^ \circ }\sin \alpha )(\sin {60^ \circ }\cos \alpha + \cos {60^ \circ }\sin \alpha ))
Now, applying the formula (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}in the above equation we get,
⇒4sin⁡α((sin⁡60∘cos⁡α−cos⁡60∘sin⁡α)(sin⁡60∘cos⁡α+cos⁡60∘sin⁡α))=4sin⁡α((sin⁡60∘cos⁡α)2−(cos⁡60∘sin⁡α)2)\Rightarrow 4\sin \alpha ((\sin {60^ \circ }\cos \alpha - \cos {60^ \circ }\sin \alpha )(\sin {60^ \circ }\cos \alpha + \cos {60^ \circ }\sin \alpha )) = 4\sin \alpha ({(\sin {60^ \circ }\cos \alpha )^2} - {(\cos {60^ \circ }\sin \alpha )^2})
Putting the values of sin⁡60∘\sin {60^ \circ } and cos⁡60∘\cos {60^ \circ }i.e. sin⁡60∘=32\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2}and cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}in the above equation we get,
⇒4sin⁡α((sin⁡60∘cos⁡α)2−(cos⁡60∘sin⁡α)2)=4sin⁡α.((32)2cos⁡2α−(12)2sin⁡2α)\Rightarrow 4\sin \alpha ({(\sin {60^ \circ }\cos \alpha )^2} - {(\cos {60^ \circ }\sin \alpha )^2}) = 4\sin \alpha .({(\dfrac{{\sqrt 3 }}{2})^2}{\cos ^2}\alpha - {(\dfrac{1}{2})^2}{\sin ^2}\alpha )
By opening the bracket of constants terms and square of them can be written as
⇒4sin⁡α.((32)2cos⁡2α−(12)2sin⁡2α)=4sin⁡α.(34cos⁡2α−14sin⁡2α)\Rightarrow 4\sin \alpha .({(\dfrac{{\sqrt 3 }}{2})^2}{\cos ^2}\alpha - {(\dfrac{1}{2})^2}{\sin ^2}\alpha ) = 4\sin \alpha .(\dfrac{3}{4}{\cos ^2}\alpha - \dfrac{1}{4}{\sin ^2}\alpha )
Taking 4 common from both terms and by cancelling it with the 4 in the numerator we get,
⇒4sin⁡α.(34cos⁡2α−14sin⁡2α)=4sin⁡α.(14(3cos⁡2α−sin⁡2α))=sin⁡α.(3cos⁡2α−sin⁡2α)\Rightarrow 4\sin \alpha .(\dfrac{3}{4}{\cos ^2}\alpha - \dfrac{1}{4}{\sin ^2}\alpha ) = 4\sin \alpha .(\dfrac{1}{4}(3{\cos ^2}\alpha - {\sin ^2}\alpha )) = \sin \alpha .(3{\cos ^2}\alpha - {\sin ^2}\alpha )
Now, by applying the formula of cos⁡2x=1−sin⁡2x{\cos ^2}x = 1 - {\sin ^2}x in the above equation we get,
⇒sin⁡α.(3cos⁡2α−sin⁡2α)=sin⁡α.(3(1−sin⁡2α)−sin⁡2α)\Rightarrow \sin \alpha .(3{\cos ^2}\alpha - {\sin ^2}\alpha ) = \sin \alpha .(3(1 - {\sin ^2}\alpha ) - {\sin ^2}\alpha )
By opening the brackets, we get,
⇒sin⁡α.(3(1−sin⁡2α)−sin⁡2α)=sin⁡α.(3−3sin⁡2α−sin⁡2α)\Rightarrow \sin \alpha .(3(1 - {\sin ^2}\alpha ) - {\sin ^2}\alpha ) = \sin \alpha .(3 - 3{\sin ^2}\alpha - {\sin ^2}\alpha )
⇒sin⁡α.(3−3sin⁡2α−sin⁡2α)=sin⁡α.(3−4sin⁡2α)\Rightarrow \sin \alpha .(3 - 3{\sin ^2}\alpha - {\sin ^2}\alpha ) = \sin \alpha .(3 - 4{\sin ^2}\alpha )
⇒sin⁡α.(3−4sin⁡2α)=3sin⁡α−4sin⁡3α\Rightarrow \sin \alpha .(3 - 4{\sin ^2}\alpha ) = 3\sin \alpha - 4{\sin ^3}\alpha
Now, from the trigonometric formula i.e. sin⁡3A=3sin⁡A−4sin⁡3A\sin 3A = 3\sin A - 4{\sin ^3}A the above equation will become i.e.
⇒3sin⁡α−4sin⁡3α=sin⁡3α\Rightarrow 3\sin \alpha - 4{\sin ^3}\alpha = \sin 3\alpha
This is equal to R.H.S. Therefore,
4sin⁡α.sin⁡(60−α).sin⁡(60+α)=sin⁡3α4\sin \alpha .\sin (60 - \alpha ).\sin (60 + \alpha ) = \sin 3\alpha L.H.S=R.H.SL.H.S = R.H.S
Hence, proved.

Note:
Here, students can get confused with terms that should be multiplied first or on which term the trigonometric formula has to be applied to obtain a result. In this first check the common variables like in this question the sine function has angle sin⁡α\sin \alpha ,sin⁡(60∘−α)\sin ({60^ \circ } - \alpha ) and sin⁡(60∘+α)\sin ({60^ \circ } + \alpha ) . You can see the last two terms have angles which are sum and difference of the same angles. Therefore, try to solve this first.
You can also do this question by another method i.e. apply other trigonometric formulae. Just remember if you are getting L.H.S is equal to R.H.S then your solution is correct either you use this method or other method.