Question

How do you prove $ \sin (a+b)+\sin (a-b)=2\sin a\cos b $ ?

Answer 1

Hint: In this question, we have to prove an equation. The equation given to us consists of trigonometric functions, so we will apply the trigonometric formula to prove the equation. We start solving the LHS of the equation and make it equal to RHS. In LHS, we will apply trigonometric formulas $ \sin (a+b)=\sin a\cos b+\cos a\sin b $ and $ \sin (a-b)=\sin a\cos b-\cos a\sin b $ in the equation. Then, we will cancel out the same terms with opposite signs and make the necessary calculations, to get the value of RHS, which is our required solution.

Complete step by step answer:
According to the question, we have to prove a trigonometric equation.
The equation is: $ \sin (a+b)+\sin (a-b)=2\sin a\cos b $ ----------- (1)
Thus, we will apply the trigonometric formula to prove equation (1).
So, we start solving this problem by taking the left-hand side of the equation.
LHS= $ \sin (a+b)+\sin (a-b) $ ----------- (2)
Now, we will apply the trigonometric identity $ \sin (a+b)=\sin a\cos b+\cos a\sin b $ and $ \sin (a-b)=\sin a\cos b-\cos a\sin b $ in equation (2), we get
 $ \Rightarrow \sin a\cos b+\cos a\sin b+\sin a\cos b-\cos a\sin b $
As we know, the same terms with opposite signs cancel out each other, therefore we get
 $ \Rightarrow \sin a\cos b+\sin a\cos b $
As we know, in addition, the same terms will add, therefore we get
 $ \Rightarrow 2\sin a\cos b=RHS $
Therefore, LHS=RHS
Hence, we proved the equation $ \sin (a+b)+\sin (a-b)=2\sin a\cos b $ .

Note:
 While proving this problem, always remember the trigonometric formula you are using and do mention it in your steps. One of the alternative methods to prove this problem is we start the question from RHS and prove it to LHS. We will apply the trigonometric formula and then make the necessary calculations to get the LHS of the equation.
An alternative method:
The equation is: $ \sin (a+b)+\sin (a-b)=2\sin a\cos b $
We start solving this problem using RHS of the above equation, that is
RHS: $ 2\sin a\cos b $ ---------- (2)
So, we will split equation (2) in terms of addition, we get
 $ \Rightarrow \sin a\cos b+\sin a\cos b $
Now, we will apply the trigonometric formula $ \sin (a+b)=\sin a\cos b+\cos a\sin b\Rightarrow \sin a\cos b=\sin (a+b)-\cos a\sin b $ and $ \sin (a-b)=\sin a\cos b-\cos a\sin b\Rightarrow \sin a\cos b=\cos (a-b)+\cos a\sin b $ in the above equation, we get
 $ \Rightarrow \sin (a+b)+\cos a\sin b+\sin (a-b)-\cos a\sin b $
As we know, the same terms with opposite signs cancel out each other, we get
 $ \Rightarrow \sin (a+b)+\sin (a-b)=LHS $
Hence proved.