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Question: How do you derive the trigonometric sum and difference formulas for sin, cos, and tan? i.e: How do I...

How do you derive the trigonometric sum and difference formulas for sin, cos, and tan? i.e: How do I derive something like sin(x+y) = sinxcosy + cosxsiny?

Explanation

Solution

To solve this problem, you can use another trigonometric identity which is cos⁡(a+b)=cos⁡acos⁡b+sin⁡asin⁡b \cos \left( a+b \right)=\cos a\cos b+\sin a\sin b. For this, first we need to convert the left hand side of the question to cos and then use the formula to get the final answer.

Complete step by step answer:
According to the problem, we are asked to derive the trigonometric sum and difference formulas for sin, cos, and tan that derive something like sin(x+y) = sinxcosy + cosxsiny.
We take the above equation as equation 1.
sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y--- ( 1 )
First we take the left hand side of the question. We convert this to cos by using the formula sin⁡(x)=cos⁡(π2−x) \sin \left( x \right)=\cos \left( \dfrac{\pi }{2}-x \right) . After converting, we get
sin⁡(x+y)=cos⁡(π2−(x+y))\sin \left( x+y \right)=\cos \left( \dfrac{\pi }{2}-\left( x+y \right) \right)
Then, applying the formula cos⁡(a+b)=cos⁡acos⁡b+sin⁡asin⁡b \cos \left( a+b \right)=\cos a\cos b+\sin a\sin b, we get
⇒cos⁡(π2−(x+y))=cos⁡((π2−x)−y)\Rightarrow \cos \left( \dfrac{\pi }{2}-\left( x+y \right) \right)=\cos \left( \left( \dfrac{\pi }{2}-x \right)-y \right)
⇒cos⁡(π2−(x+y))=cos⁡(π2−x)cos⁡y−sin⁡(π2−x)sin⁡y\Rightarrow \cos \left( \dfrac{\pi }{2}-\left( x+y \right) \right)=\cos \left( \dfrac{\pi }{2}-x \right)\cos y-\sin \left( \dfrac{\pi }{2}-x \right)\sin y ------(2)
But cos⁡(π2−x)=sin⁡x \cos \left( \dfrac{\pi }{2}-x \right) = \sin x and sin⁡(π2−x)=cos⁡x\sin \left( \dfrac{\pi }{2}-x \right) =\cos x. Therefore, after substituting these values in equation2, we get:
⇒sin⁡(x+y)=cos⁡(π2−(x+y))=sin⁡xcos⁡y+cos⁡xsin⁡y\Rightarrow \sin \left( x+y \right)=\cos \left( \dfrac{\pi }{2}-\left( x+y \right) \right)=\sin x\cos y+\cos x\sin y---Final proof
Here, you can see that the left hand side of the equation is equal to the right hand side of the equation. Hence we derived sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y \sin \left( x+y \right)=\sin x\cos y+\cos x\sin y
Therefore, we did the derivation of the given question sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y \sin \left( x+y \right)=\sin x\cos y+\cos x\sin y and proved it.

Note:
In this question, you should be careful while doing the substitutions of sin and cos. You could also derive other equations from this. If you put x=y in the given equation, you get the formula of sin2x. The above derivation can also be done by using the sin(x-b) formula.