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Trigonometric identities

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Explainer Video

Tutor: Dylan

Summary

Trigonometric identities

​​In a nutshell

Applying Pythagoras' theorem to the unit circle description of sin⁡\sinsin​, cos⁡\coscos​ and tan⁡\tantan​ allows you to deduce an important identity relation the values of sin⁡(θ)\sin(\theta)sin(θ) and cos⁡(θ)\cos(\theta)cos(θ)​ for any value of θ\thetaθ​. This, combined with the identity tan⁡(θ)≡sin⁡(θ)cos⁡(θ)\tan(\theta) \equiv \dfrac{\sin(\theta)}{\cos(\theta)}tan(θ)≡cos(θ)sin(θ)​​ (whenever the fraction is well-defined) allows you to manipulate and simplify a wide range of expressions.



Pythagoras' theorem in the unit circle

Maths; Trigonometric equations; KS5 Year 12; Trigonometric identities

​

Let PPP​ be a point on the unit circle centred at the origin such that the angle between the line OPOPOP​ and the positive xxx​-axis, measured counterclockwise from the positive xxx​-axis, is θ\thetaθ​.

Recall that the coordinates of PPP​​ are given by (cos⁡(θ),sin⁡(θ))(\cos(\theta), \sin(\theta))(cos(θ),sin(θ))​, and the gradient of the line through OOO​ and PPP​ is tan⁡(θ)\tan(\theta)tan(θ).


Note: For all values of θ\thetaθ such that cos⁡(θ)=0\cos(\theta) = 0cos(θ)=0, tan⁡(θ)\tan(\theta)tan(θ) is undefined, and it is well defined otherwise. In particular, tan⁡(θ)\tan(\theta)tan(θ) is undefined precisely whenever θ\thetaθ is an odd multiple of 90∘90^{\circ}90∘.


Pythagoras' theorem tells you that the equation for the unit circle centred at the origin is x2+y2=1x^2 + y^2 = 1x2+y2=1. Since x=cos⁡(θ)x=\cos(\theta)x=cos(θ) and y=sin⁡(θ)y=\sin(\theta)y=sin(θ), this gives the identity:


sin⁡2(θ)+cos⁡2θ≡1\boxed{\sin^2(\theta)+\cos^2{\theta}\equiv 1}sin2(θ)+cos2θ≡1​​​


Example 1

Given that sin⁡(θ)=35\sin(\theta) = \dfrac{3}{5}sin(θ)=53​, and that −90∘≤θ≤90∘-90^{\circ} \leq \theta \leq 90^{\circ}−90∘≤θ≤90∘, show that cos⁡(θ)=45\cos(\theta) = \dfrac{4}{5}cos(θ)=54​.


Substitute sin⁡(θ)=35\sin(\theta) = \dfrac35sin(θ)=53​​ into the equation sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1sin2(θ)+cos2(θ)=1, and rearrange to make cos⁡2(θ)\cos^2(\theta)cos2(θ)​ the subject:


​sin⁡2(θ)+cos⁡2(θ)=1(35)2+cos⁡2(θ)=1cos⁡2(θ)=1−925cos⁡2(θ)=1625\begin{aligned}\sin^2(\theta) + \cos^2(\theta) &= 1\\\left(\dfrac35\right)^2 + \cos^2(\theta) &= 1\\\cos^2(\theta) &= 1 - \dfrac{9}{25}\\\cos^2(\theta)&= \dfrac{16}{25}\end{aligned}sin2(θ)+cos2(θ)(53​)2+cos2(θ)cos2(θ)cos2(θ)​=1=1=1−259​=2516​​​​


Now taking the square root tells you that cos⁡(θ)=±45\cos(\theta) = \pm\dfrac45cos(θ)=±54​. However, −90∘≤θ≤90∘-90^{\circ} \leq \theta \leq 90^{\circ}−90∘≤θ≤90∘ and using a cos⁡\coscos graph, it can be seen that cos⁡(θ)≥0\cos(\theta) \geq 0cos(θ)≥0.


Therefore, cos⁡(θ)=45.‾\underline{\cos(\theta) = \dfrac45.}cos(θ)=54​.​​



An identity linking sin, cos and tan

When cos⁡(θ)≠0\cos(\theta) \neq 0cos(θ)=0​, the point (cos⁡(θ),sin⁡(θ))(\cos(\theta), \sin(\theta))(cos(θ),sin(θ))​ lies on the line y=tan⁡(θ)xy = \tan(\theta)xy=tan(θ)x, so, substituting x=cos⁡(θ)x= \cos(\theta)x=cos(θ)​ and y=sin⁡(θ)y = \sin(\theta)y=sin(θ)​:


​y=tan⁡(θ)xsin⁡(θ)=tan⁡(θ)cos⁡(θ)\begin{aligned}y &= \tan(\theta)x\\\sin(\theta) &= \tan(\theta)\cos(\theta)\end{aligned}ysin(θ)​=tan(θ)x=tan(θ)cos(θ)​​​


This is more commonly written as:

​tan⁡(θ)≡sin⁡(θ)cos⁡(θ)\boxed{\tan(\theta)\equiv \dfrac{\sin(\theta)}{\cos(\theta)}}tan(θ)≡cos(θ)sin(θ)​​​​


This holds for any value of θ\thetaθ​ for which tan⁡(θ)\tan(\theta)tan(θ)​ is well-defined​.


Example 2

Show that tan⁡2(θ)+1≡1cos⁡2(θ)\tan^2(\theta) + 1 \equiv \dfrac{1}{\cos^2(\theta)}tan2(θ)+1≡cos2(θ)1​.


Apply the identities tan⁡(θ)≡sin⁡(θ)cos⁡(θ)\tan(\theta) \equiv \dfrac{\sin(\theta)}{\cos(\theta)}tan(θ)≡cos(θ)sin(θ)​ and 1≡cos⁡2(θ)cos⁡2(θ)1 \equiv \dfrac{\cos^2(\theta)}{\cos^2(\theta)}1≡cos2(θ)cos2(θ)​ so that tan⁡2(θ)\tan^2(\theta)tan2(θ) and 111 have the same denominator:


tan⁡2(θ)+1=(sin⁡(θ)cos⁡(θ))2+cos⁡2(θ)cos⁡2(θ)=sin⁡2(θ)cos⁡2(θ)+cos⁡2(θ)cos⁡2(θ)=sin⁡2(θ)+cos⁡2(θ)cos⁡2(θ)\begin{aligned}\tan^2(\theta) +1 &= \left(\dfrac{\sin(\theta)}{\cos(\theta)}\right)^2 + \dfrac{\cos^2(\theta)}{\cos^2(\theta)}\\&= \dfrac{\sin^2(\theta)}{\cos^2(\theta)} + \dfrac{\cos^2(\theta)}{\cos^2(\theta)}\\&=\dfrac{\sin^2(\theta) + \cos^2(\theta)}{\cos^2(\theta)}\end{aligned}tan2(θ)+1​=(cos(θ)sin(θ)​)2+cos2(θ)cos2(θ)​=cos2(θ)sin2(θ)​+cos2(θ)cos2(θ)​=cos2(θ)sin2(θ)+cos2(θ)​​​​


Apply the identity sin⁡2(θ)+cos⁡2(θ)≡1\sin^2(\theta) + \cos^2(\theta) \equiv 1sin2(θ)+cos2(θ)≡1, and substitute this into the numerator:

sin⁡2(θ)+cos⁡2(θ)cos⁡2(θ)=1cos⁡2(θ)\dfrac{\sin^2(\theta) + \cos^2(\theta)}{\cos^2(\theta)} = \dfrac{1}{\cos^2(\theta)}cos2(θ)sin2(θ)+cos2(θ)​=cos2(θ)1​​​


The above holds whenever it is well-defined, therefore this is an identity.


Therefore, tan⁡2(θ)+1≡1cos⁡2(θ).‾\underline{\tan^2(\theta) +1\equiv \dfrac{1}{\cos^2(\theta)}.}tan2(θ)+1≡cos2(θ)1​.​​




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FAQs - Frequently Asked Questions

What is the most important trigonometric identity?

The most important trigonometric identity is sin^2 + cos^2 = 1.

What is the identity linking the three trigonometric functions?

The identity linking sin, cos and tan is tan(x)=sin(x)/cos(x).

What is the use of trigonometric identities?

Trigonometric identities allow you to manipulate and simplify a wide range of expressions.

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