This blog will explain how to use trigonometric identities to arrange linear and quadratic trigonometric equations into a form you can more easily solve. It will help if you’re already confident with how to solve a trigonometric equation – if you’re not, check out this blog.
If you’d like to see this blog in PDF or powerpoint formats, click here. If you need to practise some of the prior knowledge you’ll need, try these multiple-choice questions and, if you think you’ve mastered AS level trigonometry, try these exam-style questions.
An identity is a rule that’s always true, for example, (𝑎 + 𝑏)(𝑎 – 𝑏) ≡ 𝑎2 – 𝑏2 is true for any real values of 𝑎 and 𝑏. This is different to an equation, which is only true for certain values; sin𝜃 = 1 is true when 𝜃 = 90° but not when 𝜃 = 180°. There are two trigonometric identities you need to know and be able to use at AS Level:
\(\sin^2\theta +\cos^2\theta\equiv 1\)
\(\tan\theta\equiv\frac{\sin\theta}{\cos\theta}\) where \(\cos\theta\neq0\)
There are a few things to note here.
Firstly, sin2𝜃 is shorthand for (sin𝜃)2. Similarly, cos2𝜃 = (cos𝜃)2 and tan2𝜃 = (tan𝜃)2.
Secondly, the condition cos𝜃 ≠ 0 is important. If cos𝜃 = 0 then you will be dividing by 0; the result of this is undefined – this leads to the asymptotes on the graph of tan𝜃.
Both trigonometric identities can be proved using Pythagoras’ theorem and the definitions of sin, cos and tan. You need to understand these proofs.
Proof of \(\sin^2\theta+\cos^2\theta\equiv 1\)
Start by drawing a generic right-angled triangle:
You can use trigonometric ratios to show \(a=c\sin\theta\) and \(b=c\cos\theta\).
You can then substitute these into Pythagoras’ theorem:
\(\begin{aligned}&a^2+b^2=c^2\\&(c\sin\theta)^2+(c\cos\theta)^2=c^2\end{aligned}\)
Next, expand the brackets and divide by 𝑐2:
\(\begin{aligned}&c^2\sin^2\theta+c^2\cos^2\theta=c^2\\& \sin^2\theta+cos^2\theta=1\end{aligned}\)
Proof of \(\tan\theta\equiv\frac{\sin\theta}{\cos\theta}\)
Again, start by considering a generic right-angled triangle. This time you only need to work with trigonometric ratios so label the sides as O, A and H (although any labels will give the same result).
You know that \(\sin\theta=\frac{\text{O}}{\text{H}}\), \(\cos\theta=\frac{\text{A}}{\text{H}}\) and \(\tan\theta=\frac{\text{O}}{\text{A}}\). You can use that to find \(\frac{\sin\theta}{\cos\theta}\).
\(\begin{aligned}\frac{\sin\theta}{\cos\theta}
&=\frac{\text{O}}{\text{H}}\div\frac{\text{A}}{\text{H}}\\[0.5em]
&=\frac{\text{O}}{\text{H}}\times\frac{\text{H}}{\text{A}}\\[0.5em]
&=\frac{\text{OH}}{\text{HA}}\\[0.5em]
&=\frac{\text{O}}{\text{A}}\\[0.5em]
&=\tan\theta\end{aligned}\)
So, \(\tan\theta\equiv\frac{\sin\theta}{\cos\theta}\)
You will need to use these two trigonometric identities to simplify equations so that you can solve them. Let’s start by revisiting how to solve trigonometric equations.
Example Question 1
Solve \(\boldsymbol{2\cos(\frac{1}{2}x)+3=2}\) for -360° ≤ 𝑥 ≤ 360°
Start by rearranging to make \(\cos(\frac{1}{2}x)\) the subject:
\(\begin{aligned}&2\cos(\frac{1}{2}x)+3=2\\
&2\cos(\frac{1}{2}x)=-1\\
&\cos(\frac{1}{2}x)=-\frac{1}{2}\end{aligned}\)
You can then set \(\theta=\frac{1}{2}x\), or \(2\theta=x\), and substitute into the interval to give -360° ≤ 2𝜃 ≤ 360° or -180° ≤ 𝜃 ≤ 180°.
Now you can substitute into your equation and solve:
\(\begin{aligned}&\cos(\frac{1}{2}x)=-\frac{1}{2}\\
&\cos\theta=-\frac{1}{2}\\
&\theta = 120^{\circ}\end{aligned}\)
Using the symmetry of the cosine graph gives the second solution in the interval:
\(\theta = -120^{\circ}\)
Therefore, 𝑥 = 240° and -240° for -360° ≤ 𝑥 ≤ 360°.
Example Question 2
Solve \(\boldsymbol{3\sin^2\theta+5\sin\theta-2=0}\) for -360° ≤ θ ≤ 360°, giving your answers correct to 1 decimal place.
This is a quadratic equation and can be solved by using the quadratic formula, factorising or by simply using your calculator. In this case, it is a relatively simple factorisation, however you would be expected to use you calculator to solve most quadratic equations.
\(\begin{aligned}&3\sin^2\theta+5\sin\theta-2=0\\
&(3\sin\theta-1)(\sin\theta+2)=0\end{aligned}\)
Therefore, \(\sin\theta=\frac{1}{3}\) or \(\sin\theta=-2\). The second solution is not possible as \(\sin\theta\) is only defined between 1 and -1. Solve \(\sin\theta=\frac{1}{3}\):
\(\sin\theta=\frac{1}{3}\)
θ = -340.5°, -199.5°, 19.5° and 160.5° for -360° ≤ θ ≤ 360°
Example Question 3
Solve \(\boldsymbol{2\sin(3x)=3\cos(3x)}\) for 0° ≤ 𝑥 ≤ 180°, giving your answers correct to 1 decimal place.
For this equation, you need to use \(\tan\theta\equiv\frac{\sin\theta}{\cos\theta}\), as the equation involves both sine and cosine and doesn’t have any squared terms. As you practise more questions of this type, you’ll start to recognise which of the trigonometric identities to use.
Start by dividing both sides by \(\cos(3x)\):
\(\begin{aligned}&2\sin(3x)=3\cos(3x)\\
&\frac{2\sin(3x)}{\cos(3x)}=3\end{aligned}\)
You can now replace \(\frac{\sin(3x)}{\cos(3x)}\) with \(\tan(3x)\) and solve the equation:
\(\begin{aligned}&2\tan(3x)=3\end{aligned}\)
You need to solve for 0° ≤ 𝑥 ≤ 180°. Set θ = 3𝑥, or 𝑥 = \(\frac{\theta}{3}\). This means that 0° ≤ θ ≤ 540°.
\(\begin{aligned}&2\tan\theta=3\\
&\tan\theta=\frac{3}{2}\\
&\theta = 56.3^{\circ}, 236.3^{\circ} \text{ and }416.3^{\circ}\text{ for }0^{\circ}\leq\theta\leq 540^{\circ}\end{aligned}\)
Therefore, 𝑥 = 18.8°, 78.8° and 138.8° for 0° ≤ 𝑥 ≤ 180°.
Example Question 4
Solve \(\boldsymbol{8-3\cos\theta=9\sin^2\theta}\) for 0° ≤ \(\boldsymbol{\theta}\) ≤ 360°, giving your answers correct to 1 decimal place.
This time, you need to use a version of \(\sin^2\theta+\cos^2\theta\equiv 1\) to convert the equation to a single trigonometric ratio, as the equation contains the term \(\sin^2\theta\). You can rearrange the identity to give \(\sin^2\theta\equiv 1-\cos^2\theta\).
\(\begin{aligned}&8-3\cos\theta=9\sin^2\theta\\
&8-3\cos\theta=9(1-\cos^2\theta)\\
&8-3\cos\theta=9-9\cos^2\theta\\
&8-3\cos\theta+9\cos^2\theta=9\\
&9\cos^2\theta-3\cos\theta-1=0\end{aligned}\)
Substitute \(x=\cos\theta\) and use your calculator to solve the quadratic \(9x^2-3x-1\). Doing so gives two answers:
\(x=\frac{1+\sqrt{5}}{6}=0.539…\) and \(x=\frac{1-\sqrt{5}}{6}=-0.206…\)
Always use the exact values to continue solving, but it’s also important to consider the decimal value to check if the solution is between -1 and 1.
You now have two trigonometric equations to solve, \(\cos\theta=\frac{1+\sqrt{5}}{6}\) and \(\cos\theta=\frac{1-\sqrt{5}}{6}\):
\(\begin{aligned}&\cos\theta=\frac{1+\sqrt{5}}{6}\\&\theta=\cos^{-1}(\frac{1+\sqrt{5}}{6})\\&\theta=57.4^{\circ}\text{ and }302.6^{\circ}\end{aligned}\)
\(\begin{aligned}&\cos\theta=\frac{1-\sqrt{5}}{6}\\&\theta=\cos^{-1}(\frac{1-\sqrt{5}}{6})\\&\theta=101.9^{\circ}\text{ and }258.1^{\circ}\end{aligned}\)
The final four answers for 0° ≤ θ ≤ 360° are:
θ = 57.4°, 101.9°, 258.1° and 302.6°
Practice Questions
1. Solve \(3\cos^2\theta+2\cos\theta-4=0\) for -180° ≤ θ ≤ 180°.
2. Solve \(\sin^2\theta-\sin\theta + 14 = 0\) for -180° ≤ θ ≤ 180°.
3. Solve \(6\tan^2\theta-7\tan\theta-3=0\) for -180° ≤ θ ≤ 180°.
4. Solve \(\cos^2x=3\sin{x}-1\) for 0° ≤ 𝑥 ≤ 180°.
5. Solve \(4\sin\theta=2\cos\theta\) for 0° ≤ θ ≤ 360°.
6. Solve \(4\sin(3x)=4\cos(3x)\) for 0° ≤ 𝑥 ≤ 180°.
7. Solve \(\cos^2(\theta+30)+3=4\sin^2(\theta+30)-\sin(\theta+30)\) for 0° ≤ θ ≤ 360°.
8. Solve \(\sin{x}=\frac{3\cos{x}\,-\,2}{\tan{x}}\) for -180° ≤ 𝑥 ≤ 180°.
Answers
Did that answer your burning questions on trigonometric identities? We have a great range of helpful blogs here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free resources before you subscribe too.