Basic Trig Identities

A LevelAQAEdexcelOCR

Basic Trig Identities

We’ve still got one or two more rules to know…

These ones relate sin⁡textcolor{blue}{sin}, cos⁡textcolor{limegreen}{cos} and tan⁡textcolor{red}{tan} together.

A LevelAQAEdexcelOCR
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Rule 1

tan⁡x≡sin⁡xcos⁡xtextcolor{red}{tan x} equiv dfrac{textcolor{blue}{sin x}}{textcolor{limegreen}{cos x}}

Don’t be threatened by the strange equals sign here – that’s the notation we use for an identity. This expression is true for every value of xx.

This is really easy to prove:

sin⁡x=oppositehypotenusetextcolor{blue}{sin x} = dfrac{text{opposite}}{text{hypotenuse}} and cos⁡x=adjacenthypotenusetextcolor{limegreen}{cos x} = dfrac{text{adjacent}}{text{hypotenuse}}

Then, sin⁡xcos⁡x=(oppositehypotenuse)(adjacenthypotenuse)=(oppositehypotenuse)(adjacenthypotenuse)=oppositeadjacent=tan⁡xdfrac{textcolor{blue}{sin x}}{textcolor{limegreen}{cos x}} = dfrac{left( dfrac{text{opposite}}{text{hypotenuse}}right) }{left( dfrac{text{adjacent}}{text{hypotenuse}}right) } = dfrac{left( dfrac{text{opposite}}{cancel{text{hypotenuse}}}right) }{left( dfrac{text{adjacent}}{cancel{text{hypotenuse}}}right) } = dfrac{text{opposite}}{text{adjacent}} = textcolor{red}{tan x}

Messy, but easy.

A LevelAQAEdexcelOCR

Rule 2

sin⁡2x+cos⁡2x≡1textcolor{blue}{sin ^2 x} + textcolor{limegreen}{cos ^2 x} equiv 1

Actually, this identity is an extension of Pythagoras’ Theorem.

Think about the sin⁡textcolor{blue}{sin} and cos⁡textcolor{limegreen}{cos} functions.

In this identity, we’re suggesting that (oppositehypotenuse)2+(adjacenthypotenuse)2=1bigg(dfrac{text{opposite}}{text{hypotenuse}}bigg) ^2 + bigg(dfrac{text{adjacent}}{text{hypotenuse}}bigg) ^2 = 1.

Well, let’s denote the opposite as aa, the adjacent as bb, and the hypotenuse as cc.

From here, we have a2c2+b2c2=1dfrac{a^2}{c^2} + dfrac{b^2}{c^2} = 1

We can express 11 as c2c2dfrac{c^2}{c^2}, which gives a2c2+b2c2=c2c2dfrac{a^2}{c^2} + dfrac{b^2}{c^2} = dfrac{c^2}{c^2}, or a2+b2=c2a^2 + b^2 = c^2.

A LevelAQAEdexcelOCR
A LevelAQAEdexcelOCR

Example: Adapting Equations

Find the values of 0≤x≤2π0 leq x leq 2pi, such that cos⁡x−tan⁡x=0textcolor{limegreen}{cos x} – textcolor{red}{tan x} = 0.

[4 marks]

So, first things first, let’s use our first identity to get

cos⁡x−sin⁡xcos⁡x=0textcolor{limegreen}{cos x} – dfrac{textcolor{blue}{sin x}}{textcolor{limegreen}{cos x}} = 0

Now, we can multiply up by cos⁡xtextcolor{limegreen}{cos x} on both sides to get

cos⁡2x−sin⁡x=0textcolor{limegreen}{cos ^2 x} – textcolor{blue}{sin x} = 0

Now, use the second identity to give

1−sin⁡2x−sin⁡x=01 – textcolor{blue}{sin ^2 x} – textcolor{blue}{sin x} = 0

or

sin⁡2x+sin⁡x−1=0textcolor{blue}{sin ^ 2 x} + textcolor{blue}{sin x} – 1 = 0

This looks a little like a quadratic equation… Let y=sin⁡xy = textcolor{blue}{sin x}:

y2+y−1=0y^2 + y – 1 = 0

which has solutions

y=−1±12−(4×1×−1)2=−1±52y = dfrac{-1 ± sqrt{1^{2} – (4 times 1 times -1)}}{2} = dfrac{-1 ± sqrt{5}}{2}

Since

sin⁡x≠−1−52textcolor{blue}{sin x} neq dfrac{-1 – sqrt{5}}{2} for any xx, we have sin⁡x=−1+52textcolor{blue}{sin x} = dfrac{-1 + sqrt{5}}{2}

which gives

x=sin⁡−1(−1+52)=0.666,2.475x = sin ^{-1}left( dfrac{-1 +sqrt{5}}{2}right) = 0.666, 2.475

A LevelAQAEdexcelOCR

Basic Trig Identities Example Questions

Question 1: Find the solutions for 2sin⁡x−tan⁡x=02sin x – tan x = 0 in the interval 0≤x≤2π0 leq x leq 2pi.

[4 marks]

A Level AQAEdexcelOCR

2sin⁡x−tan⁡x=02sin x – tan x = 0

is equivalent to

2sin⁡x−sin⁡xcos⁡x=02sin x – dfrac{sin x}{cos x} = 0

or

2sin⁡xcos⁡x−sin⁡x=02sin x cos x – sin x = 0

which can be factorised to

(2cos⁡x−1)sin⁡x=0(2cos x – 1)sin x = 0

so

sin⁡x=0sin x = 0 or cos⁡x=12cos x = dfrac{1}{2}

 

sin⁡x=0sin x = 0 has solutions at x=0,π,2πx = 0, pi, 2pi

and

cos⁡x=12cos x = dfrac{1}{2} has solutions at x=π3,5π3x = dfrac{pi}{3}, dfrac{5pi}{3}

So, there are five solutions, x=0,π3,π,5π3,2πx = 0, dfrac{pi}{3}, pi, dfrac{5pi}{3}, 2pi.

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Question 2: Show that (sin⁡x+cos⁡x)(sin⁡x−cos⁡x)=0(sin x + cos x)(sin x – cos x) = 0 has solutions x=45°,135°,225°,315°x = 45°, 135°, 225°, 315° for the interval 0°≤x≤360°0° leq x leq 360°.

[3 marks]

A Level AQAEdexcelOCR

(sin⁡x+cos⁡x)(sin⁡x−cos⁡x)(sin x + cos x)(sin x – cos x)

=sin⁡2x−cos⁡2x= sin ^2 x – cos ^2 x

=2sin⁡2x−1= 2sin ^2 x – 1

so we require

sin⁡x=±12sin x = ±dfrac{1}{sqrt{2}}

which gives us the set of solutions

x=45°,135°,225°,315°x = 45°, 135°, 225°, 315°

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Question 3: Explain briefly why tan⁡xtan x has no value when x=90°,270°,450°,...,90°±180°nx = 90°, 270°, 450°, …, 90° ± 180°n.

[3 marks]

A Level AQAEdexcelOCR

Use the first identity, tan⁡x≡sin⁡xcos⁡xtan x equiv dfrac{sin x}{cos x}

For each value of x=90°,270°,450°,...,90°±180°nx = 90°, 270°, 450°, …, 90° ± 180°n, we have cos⁡x=0cos x = 0.

Since we cannot divide by 00, we must have tan⁡xtan x undefined at these values.

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Specification Points Covered

E5 – Understand and use tan⁡θ=sin⁡θcos⁡θtan{theta}=dfrac{sin{theta}}{cos{theta}}
Understand and use sin⁡2θ+cos⁡2θ=1sin^2{theta}+cos^2{theta}=1; sec⁡2θ=1+tan⁡2θsec^2{theta}=1+tan^2{theta} and cosec⁡2θ=1+cot⁡2θcosec^2{theta}=1+cot^2{theta}

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