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Question: How do you derive the multiple angles formula?...

How do you derive the multiple angles formula?

Explanation

Solution

To solve this question, you could use the trigonometric identities or also you could use the euler's theorem. If you use trigonometric identities, you have to substitute all the different values with the same value. Then you can derive the formulas.

Complete step by step answer:
According to the problem, we are asked to derive the multiple angles formula.
Multiple angles formula is the formulas for double angle, triple angle and more. In this problem, we are going to derive the double angle formula for sin. Therefore, using the same method, we can derive the multiple angle formulas for sin, cos, tan, cot, sec, and cosec.
To do this, first I use the trigonometric identity if sine, which is:
sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y--- ( 1 )
And we can consider this as equation 1.
Now, in this we should substitute y as x, that is x = y. Substituting this in equation 1, we get
⇒sin⁡(x+x)=sin⁡xcos⁡x+cos⁡xsin⁡x\Rightarrow \sin \left( x+x \right)=\sin x\cos x+\cos x\sin x
⇒sin⁡(2x)=2sin⁡xcos⁡x\Rightarrow \sin \left( 2x \right)=2\sin x\cos x---Final answer
Here, as you can see that, we derived the double angle formula for the sine function, that issin⁡(2x)=2sin⁡xcos⁡x \sin \left( 2x \right)=2\sin x\cos x. For triple angle we use again the same formula sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y \sin \left( x+y \right)=\sin x\cos y+\cos x\sin y , but here, we substitute x as 2x and y as x.
⇒sin⁡(2x+x)=sin⁡2xcos⁡x+cos⁡2xsin⁡x\Rightarrow \sin \left( 2x+x \right)=\sin 2x\cos x+\cos 2x\sin x
⇒sin⁡(3x)=(2sin⁡xcos⁡x)cos⁡x+(1−2sin⁡2x)sin⁡x\Rightarrow \sin \left( 3x \right)=\left( 2\sin x\cos x \right)\cos x+\left( 1-2{{\sin }^{2}}x \right)\sin x
⇒sin⁡(3x)=2sin⁡xcos⁡2x−2sin⁡3x+sin⁡x\Rightarrow \sin \left( 3x \right)=2\sin x{{\cos }^{2}}x-2{{\sin }^{3}}x+\sin x
⇒sin⁡(3x)=2sin⁡x(1−sin⁡2x)−2sin⁡3x+sin⁡x\Rightarrow \sin \left( 3x \right)=2\sin x\left( 1-{{\sin }^{2}}x \right)-2{{\sin }^{3}}x+\sin x
⇒sin⁡(3x)=2sin⁡x−2sin⁡3x−2sin⁡3x+sin⁡x\Rightarrow \sin \left( 3x \right)=2\sin x-2{{\sin }^{3}}x-2{{\sin }^{3}}x+\sin x
⇒sin⁡(3x)=3sin⁡x−4sin⁡3x\Rightarrow \sin \left( 3x \right)=3\sin x-4{{\sin }^{3}}x
Therefore, we got the triple formula for sine.
Similarly, we can derive the multiple angle formulas for sin, cos, tan, cot, sec, and cosec.
Therefore, we did the derivation of multiple angles formulas.

Note:
While doing this question, we should be careful in doing the substitutions of cos or sin or tan. Also, you should try to remember all the trigonometric identities so that you can solve the trigonometric questions easily. We could also derive the multiple angles formula using the euler’s formula.