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Question: How do you verify the identity \(\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos y\) ?...

How do you verify the identity sin⁡(x+y)+sin⁡(x−y)=2sin⁡xcos⁡y\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos y ?

Explanation

Solution

We are asked to verify that sin⁡(x+y)+sin⁡(x−y)=2sin⁡xcos⁡y\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos y.
To solve this we will first learn how sin behaves when applied on the sum of two numbers and on the difference of two numbers.
We will use sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y and
We will use sin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡y\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y
We start out by considering the left hand side part of the problem, we use the above formula and simplify and will record to the right side of the given equation.

Complete step by step answer:
We are given that we have to verify that sin⁡(x+y)+sin⁡(x−y)=2sin⁡xcos⁡y\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos y.
To do that solution we will need to know how one will be able to use the function sin on the sum of two numbers and also on the difference of two numbers.
We have to learn that what does sin⁡(x+y)\sin \left( x+y \right) expand into and we will also know the knowledge that how does the sin⁡(x−y)\sin \left( x-y \right) will expand itself into .
We know that –
sin⁡(A+B)\sin \left( A+B \right) is given as sin⁡Acos⁡B+cos⁡A+sin⁡B\sin A\cos B+\cos A+\sin B
While sin⁡(A−B)\sin \left( A-B \right) is given as sin⁡Acos⁡B−cos⁡Asin⁡B\sin A\cos B-\cos A\sin B
We consider the left hand side.
Now we are asked to find the value of sin⁡(x+y)+sin⁡(x−y)\sin \left( x+y \right)+\sin \left( x-y \right)
So,
Using above formula
Considering A as x and B as y, we get –
sin⁡(x+y)+sin⁡(x−y)\sin \left( x+y \right)+\sin \left( x-y \right) will become
sin⁡(x+y)+sin⁡(x−y)=sin⁡xcos⁡y+cos⁡xsin⁡y+sin⁡xcos⁡y−cos⁡xsin⁡y\sin \left( x+y \right)+\sin \left( x-y \right)=\sin x\cos y+\cos x\sin y+\sin x\cos y-\cos x\sin y
We can see that –
+cos⁡xsin⁡y−cos⁡xsin⁡y=0+\cos x\sin y-\cos x\sin y=0 , so we get –
sin⁡(x+y)+sin⁡(x−y)=sin⁡xcos⁡y+sin⁡xcos⁡y+0 =sin⁡xcos⁡y+sin⁡xcos⁡y \begin{aligned} & \sin \left( x+y \right)+\sin \left( x-y \right)=\sin x\cos y+\sin x\cos y+0 \\\ & =\sin x\cos y+\sin x\cos y \\\ \end{aligned}
Now, as sin⁡xcos⁡y\sin x\cos y adding two times it become 2sin⁡xcos⁡y2\sin x\cos y ,
So, we get –
sin⁡(x+y)+sin⁡(x−y)=2sin⁡xcos⁡y\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos y
We got the right hand side so we got our answer.
L.H.S=R.H.S
Hence proved.

Note: We can cross check this identity
sin⁡(x+y)+sin⁡(x−y)=2sin⁡xcos⁡y\sin \left( x+y \right)+\sin \left( x-y \right)=2\sin x\cos yis valid
For any x and y
Let x=60∘x={{60}^{\circ }} and y=30∘y={{30}^{\circ }}
So, x+y=60∘+30∘=90∘x+y={{60}^{\circ }}+{{30}^{\circ }}={{90}^{\circ }}
And x−y=60∘−30∘=30∘x-y={{60}^{\circ }}-{{30}^{\circ }}={{30}^{\circ }}
So,
First left hand side value
sin⁡(x+y)+sin⁡(x−y)=sin⁡(90∘)+sin⁡(30∘)\sin \left( x+y \right)+\sin \left( x-y \right)=\sin \left( {{90}^{\circ }} \right)+\sin \left( {{30}^{\circ }} \right)
As sin⁡90∘=1\sin {{90}^{\circ }}=1 and sin⁡30∘=12\sin {{30}^{\circ }}=\dfrac{1}{2} So we get –
sin⁡(x+y)+sin⁡(x−y)=1+12=32\sin \left( x+y \right)+\sin \left( x-y \right)=1+\dfrac{1}{2}=\dfrac{3}{2}
Now considering right hand side
Putting x=60∘x={{60}^{\circ }} and y=30∘y={{30}^{\circ }} , we get –
2sin⁡xcos⁡y=2×sin⁡60∘×cos⁡30∘2\sin x\cos y=2\times \sin {{60}^{\circ }}\times \cos {{30}^{\circ }}
As sin⁡60∘=32\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} and cos⁡30∘=32\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}
So,
2sin⁡xcos⁡y=2×32×322\sin x\cos y=2\times \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}
By simplifying, we get –
=2×3×34=\dfrac{2\times \sqrt{3}\times \sqrt{3}}{4}
As 3×3=3\sqrt{3}\times \sqrt{3}=3 so we get –
2sin⁡xcos⁡y=2×34=322\sin x\cos y=\dfrac{2\times 3}{4}=\dfrac{3}{2} ,
So, we get –
L.H.S=R.H.S
Hence, this identity stands true.