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Exact trigonometric values

Exact trigonometric values

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Summary

Exact trigonometric values

​​In a nutshell

It is possible to compute the exact values of sin⁡\sinsin​, cos⁡\coscos​ and tan⁡\tantan​ for the angles 30∘30^{\circ}30∘​, 45∘45^{\circ}45∘​ and 60∘60^{\circ}60∘​ by considering an equilateral triangle and an isoceles right-angled triangle.



Exact values from the equilateral triangle ​​

To compute sin⁡\sinsin​, cos⁡\coscos​ and tan⁡\tantan​ of 30∘30^{\circ}30∘ and 60∘60^{\circ}60∘​, consider an equilateral triangle of side length 111, and draw a line perpendicular to one of the sides from its midpoint to the opposite corner.

Maths; Trigonometric equations; KS5 Year 12; Exact trigonometric values

The perpendicular bisects the triangle into two congruent right-angled triangles, with angles 30∘30^{\circ}30∘​, 60∘60^{\circ}60∘​ and 90∘90^{\circ}90∘. The hypotenuse has length 111​, and one of the sides has length 12\dfrac{1}{2}21​. Call the length of the third side ​ccc. Using Pythagoras' theorem to compute ccc​:


​12=(12)2+c21=14+c234=c2±32=c\begin{aligned}1^2 &= (\dfrac{1}{2})^2 + c^2 \\1 &= \dfrac{1}{4} + c^2 \\\dfrac{3}{4} &= c^2 \\\pm\dfrac{\sqrt{3}}{2} &= c\end{aligned}12143​±23​​​=(21​)2+c2=41​+c2=c2=c​​​


However ccc​ is a length, therefore positive, so deduce that c=32c = \dfrac{\sqrt{3}}{2}c=23​​​.


Trigonometric ratios (SOHCAHTOA) in the right-angled triangle with side lengths 111​, 12\dfrac{1}{2}21​​ and 32\dfrac{\sqrt{3}}{2}23​​ allow you to compute:


​sin⁡(30∘)=12cos⁡(30∘)=32tan⁡(30∘)=13=33sin⁡(60∘)=32cos⁡(60∘)=12tan⁡(60∘)=3\boxed{\begin{aligned}\sin(30^{\circ}) &= \dfrac{1}{2}\\\cos(30^{\circ}) &= \dfrac{\sqrt{3}}{2}\\\tan(30^{\circ}) &= \dfrac{1}{\sqrt{3}}= \dfrac{\sqrt{3}}{3}\\\\\sin(60^{\circ}) &= \dfrac{\sqrt{3}}{2}\\\cos(60^{\circ}) &= \dfrac{1}{2}\\\tan(60^{\circ}) &= \sqrt{3}\end{aligned}}sin(30∘)cos(30∘)tan(30∘)sin(60∘)cos(60∘)tan(60∘)​=21​=23​​=3​1​=33​​=23​​=21​=3​​​​​


​​Example 1

What is the exact value of sin⁡(150∘)\sin(150^{\circ})sin(150∘)?


Express sin⁡(150∘)\sin(150^{\circ})sin(150∘) in terms of the acute angle 30∘30^{\circ}30∘ by substituting θ=30∘\theta = 30^{\circ}θ=30∘ into the identity sin⁡(180∘−θ)=sin⁡(θ)\sin(180^{\circ} - \theta) = \sin(\theta)sin(180∘−θ)=sin(θ):


sin⁡(180∘−θ)=sin⁡(θ)sin⁡(180∘−30∘)=sin⁡(30∘)sin⁡(150∘)=sin⁡(30∘)\begin{aligned}\sin(180^{\circ} - \theta) &= \sin(\theta)\\\sin(180^{\circ} - 30^{\circ}) &= \sin(30^{\circ}) \\\sin(150^{\circ}) &= \sin(30^{\circ})\end{aligned}sin(180∘−θ)sin(180∘−30∘)sin(150∘)​=sin(θ)=sin(30∘)=sin(30∘)​​​

Now use the fact that sin⁡(30∘)=12\sin(30^{\circ}) = \dfrac{1}{2}sin(30∘)=21​ to deduce that sin⁡(150∘)=12\sin(150^{\circ}) = \dfrac12sin(150∘)=21​​


sin⁡(150∘)=12.‾\underline{\sin(150^{\circ}) = \dfrac12.}sin(150∘)=21​.​​​

​​


Exact values from the isosceles right-angled triangle

To compute sin⁡\sinsin​, cos⁡\coscos​ and tan⁡\tantan​ of 45∘45^{\circ}45∘, consider an isosceles right-angled triangle with a hypotenuse of length 111. 

Maths; Trigonometric equations; KS5 Year 12; Exact trigonometric values

This triangle has two 45∘45^{\circ}45∘​ angles, and one right angle. Call the length of the two shorter sides aaa​, and use Pythagoras' theorem to find aaa​:


12=a2+a21=2a212=a2±12=a±22=a\begin{aligned}1^2 &= a^2 + a^2\\1 &= 2a^2\\\dfrac{1}{2} &= a^2\\\pm\dfrac{1}{\sqrt{2}}&=a\\\pm\dfrac{\sqrt{2}}{2}&=a\end{aligned}12121​±2​1​±22​​​=a2+a2=2a2=a2=a=a​​​


However aaa​ is a length, so must be positive, hence deduce that a=22a = \dfrac{\sqrt{2}}{2}a=22​​​.


Trigonometric ratios (SOHCAHTOA) in the isosceles right-angled triangle with a hypotenuse of length 111 allow you to compute:


sin⁡(45∘)=22cos⁡(45∘)=22tan⁡(45∘)=1\boxed{\begin{aligned}\sin(45^{\circ}) &= \dfrac{\sqrt{2}}{2}\\\cos(45^{\circ}) &= \dfrac{\sqrt{2}}{2}\\\tan(45^{\circ})&=1\end{aligned} }sin(45∘)cos(45∘)tan(45∘)​=22​​=22​​=1​​​


Example 2

What is the exact value of tan⁡(315∘)\tan(315^{\circ})tan(315∘)?


Express tan⁡(315∘)\tan(315^{\circ})tan(315∘) in terms of the acute angle 45∘45^{\circ}45∘ by substituting θ=45∘\theta = 45^{\circ}θ=45∘ into the identity tan⁡(360∘−θ)=−tan⁡(θ)\tan(360^{\circ}- \theta) = -\tan(\theta)tan(360∘−θ)=−tan(θ):


tan⁡(360∘−θ)=−tan⁡(θ)tan⁡(360∘−45∘)=−tan⁡(45∘)tan⁡(315∘)=−tan⁡(45∘)\begin{aligned}\tan(360^{\circ} - \theta) &= -\tan(\theta)\\\tan(360^{\circ} - 45^{\circ}) &= -\tan(45^{\circ})\\\tan(315^{\circ}) &= -\tan(45^{\circ})\end{aligned}tan(360∘−θ)tan(360∘−45∘)tan(315∘)​=−tan(θ)=−tan(45∘)=−tan(45∘)​​​


Now use the fact that tan⁡(45∘)=1\tan(45^{\circ}) = 1tan(45∘)=1 to deduce that tan⁡(315∘)=−1\tan(315^{\circ}) = -1tan(315∘)=−1​


tan⁡(315∘)=−1.‾\underline{\tan(315^{\circ}) = -1.} tan(315∘)=−1.​






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FAQs - Frequently Asked Questions

What is the exact value of sin30?

The exact value of sin30 is 1/2.

What is the exact value of tan30?

The exact value of tan30 is √3/3.

How do I find the exact value of trigonometric ratios?

It is possible to compute the exact values of sin, cos and tan for the angles 30, 45 and 90 by considering an equilateral triangle and an isoceles right-angled triangle.

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