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Proving trigonometric identities

Proving trigonometric identities

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Explainer Video

Tutor: Toby

Summary

Proving trigonometric identities

​​In a nutshell

You can use all the trigonometric identities that you have learned so far to prove more trigonometric identities.


Proving trigonometric identities

Proofs may involve using the reciprocal trigonometric functions, the addition formulae and double angle formulae.


Example 1

Prove that

​tan⁡x+cot⁡x≡2cosec⁡2x\tan x + \cot x \equiv 2 \cosec 2xtanx+cotx≡2cosec2x​​


Start from the left hand side and change all trig functions into sin⁡\sinsin and cos⁡\coscos:

tan⁡x+cot⁡x=sin⁡xcos⁡x+cos⁡xsin⁡x=sin⁡2x+cos⁡2xsin⁡xcos⁡x \begin{aligned}\tan x + \cot x &= \dfrac{\sin x}{\cos x} +\dfrac{\cos x}{\sin x} \\\\&= \dfrac{\sin^2x + \cos^2x}{\sin x \cos x} \\\ \end{aligned}tanx+cotx ​=cosxsinx​+sinxcosx​=sinxcosxsin2x+cos2x​​​​


Use sin⁡2x+cos⁡2x=1\sin^2x + \cos^2x=1sin2x+cos2x=1​

​sin⁡2x+cos⁡2xsin⁡xcos⁡x=1sin⁡xcos⁡x=22sin⁡xcos⁡x\begin{aligned} \dfrac{\sin^2x + \cos^2x}{\sin x \cos x} & = \dfrac{1}{\sin x \cos x} \\\\& = \dfrac{2}{2 \sin x \cos x} \\ \end{aligned}sinxcosxsin2x+cos2x​​=sinxcosx1​=2sinxcosx2​​​​


Use the double angle formula sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos xsin2x=2sinxcosx:

​22sin⁡xcos⁡x=2sin⁡2x=2cosec⁡2x\begin{aligned} \dfrac{2}{2 \sin x \cos x} & = \dfrac{2}{ \sin 2x } \\\\&= 2 \cosec 2x\end{aligned}2sinxcosx2​​=sin2x2​=2cosec2x​​​


Therefore, tan⁡x+cot⁡x≡2cosec⁡2x‾\underline{\tan x + \cot x \equiv 2 \cosec 2x}tanx+cotx≡2cosec2x​.


Example 2

Prove that

​cos⁡3x≡4cos⁡3x−3cos⁡x\cos 3x \equiv 4 \cos^3 x - 3 \cos xcos3x≡4cos3x−3cosx​​


Start from the left hand side and write cos⁡3x\cos 3xcos3x as cos⁡(2x+x)\cos (2x+x)cos(2x+x) and use the addition formula:

cos⁡3x=cos⁡(2x+x)=cos⁡2xcos⁡x−sin⁡2xsin⁡x\begin{aligned}\cos 3x &= \cos (2x+x) \\&= \cos 2x \cos x - \sin 2x \sin x \\\end{aligned}cos3x​=cos(2x+x)=cos2xcosx−sin2xsinx​​​


Use the double angle formulae to give:

​cos⁡2xcos⁡x−sin⁡2xsin⁡x=(2cos⁡2x−1)cos⁡x−2sin⁡xcos⁡xsin⁡x=2cos⁡3x−cos⁡x−2cos⁡xsin⁡2x\begin{aligned} \cos 2x \cos x - \sin 2x \sin x &= (2\cos^2x-1) \cos x - 2\sin x \cos x \sin x \\&= 2 \cos^3 x - \cos x - 2 \cos x \sin^2 x \\\end{aligned}cos2xcosx−sin2xsinx​=(2cos2x−1)cosx−2sinxcosxsinx=2cos3x−cosx−2cosxsin2x​​​


Use cos⁡2x+sin⁡2x=1\cos^2x + \sin^2 x=1cos2x+sin2x=1 to convert to cos⁡\coscos:

​2cos⁡3x−cos⁡x−2cos⁡xsin⁡2x=2cos⁡3x−cos⁡x−2cos⁡x(1−cos⁡2x)=2cos⁡3x−cos⁡x−2cos⁡x+2cos⁡3x=4cos⁡3x−3cos⁡x\begin{aligned} 2 \cos^3 x - \cos x - 2 \cos x \sin^2 x &= 2 \cos^3 x - \cos x - 2 \cos x (1- \cos^2 x) \\& = 2 \cos^3x - \cos x - 2 \cos x + 2 \cos^3 x \\&= 4 \cos^3 x - 3 \cos x \end{aligned}2cos3x−cosx−2cosxsin2x​=2cos3x−cosx−2cosx(1−cos2x)=2cos3x−cosx−2cosx+2cos3x=4cos3x−3cosx​​​


Therefore, cos⁡3x≡4cos⁡3x−3cos⁡x‾\underline{\cos 3x \equiv 4 \cos^3 x - 3 \cos x}cos3x≡4cos3x−3cosx​.




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Equations and identities

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Equations and identities

Trigonometric identities

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Trigonometric identities

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Proving trigonometric identities

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FAQs - Frequently Asked Questions

How do you prove a trig identity?

Start from either side and use other trig identities you have already learned to prove that you can obtain the other side of the identitiy.

What kind of identities do you need to know in order to prove new trig identities?

Proves may involve using the reciprocal trigonometric functions, the addition formulae and double angle formulae.

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