# Basic Trig Identities

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Basic Trig Identities

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

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

A Level   ## Rule 1

$\textcolor{red}{\tan x} \equiv \dfrac{\textcolor{blue}{\sin x}}{\textcolor{limegreen}{\cos x}}$

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 $x$.

This is really easy to prove:

$\textcolor{blue}{\sin x} = \dfrac{\text{opposite}}{\text{hypotenuse}}$ and $\textcolor{limegreen}{\cos x} = \dfrac{\text{adjacent}}{\text{hypotenuse}}$

Then, $\dfrac{\textcolor{blue}{\sin x}}{\textcolor{limegreen}{\cos x}} = \dfrac{\left( \dfrac{\text{opposite}}{\text{hypotenuse}}\right) }{\left( \dfrac{\text{adjacent}}{\text{hypotenuse}}\right) } = \dfrac{\left( \dfrac{\text{opposite}}{\cancel{\text{hypotenuse}}}\right) }{\left( \dfrac{\text{adjacent}}{\cancel{\text{hypotenuse}}}\right) } = \dfrac{\text{opposite}}{\text{adjacent}} = \textcolor{red}{\tan x}$

Messy, but easy.

A Level   ## Rule 2

$\textcolor{blue}{\sin ^2 x} + \textcolor{limegreen}{\cos ^2 x} \equiv 1$

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

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

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

Well, let’s denote the opposite as $a$, the adjacent as $b$, and the hypotenuse as $c$.

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

We can express $1$ as $\dfrac{c^2}{c^2}$, which gives $\dfrac{a^2}{c^2} + \dfrac{b^2}{c^2} = \dfrac{c^2}{c^2}$, or $a^2 + b^2 = c^2$.

A Level   A Level   Find the values of $0 \leq x \leq 2\pi$, such that $\textcolor{limegreen}{\cos x} - \textcolor{red}{\tan x} = 0$.

[4 marks]

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

$\textcolor{limegreen}{\cos x} - \dfrac{\textcolor{blue}{\sin x}}{\textcolor{limegreen}{\cos x}} = 0$

Now, we can multiply up by $\textcolor{limegreen}{\cos x}$ on both sides to get

$\textcolor{limegreen}{\cos ^2 x} - \textcolor{blue}{\sin x} = 0$

Now, use the second identity to give

$1 - \textcolor{blue}{\sin ^2 x} - \textcolor{blue}{\sin x} = 0$

or

$\textcolor{blue}{\sin ^ 2 x} + \textcolor{blue}{\sin x} - 1 = 0$

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

$y^2 + y - 1 = 0$

which has solutions

$y = \dfrac{-1 ± \sqrt{1^{2} - (4 \times 1 \times -1)}}{2} = \dfrac{-1 ± \sqrt{5}}{2}$

Since

$\textcolor{blue}{\sin x} \neq \dfrac{-1 - \sqrt{5}}{2}$ for any $x$, we have $\textcolor{blue}{\sin x} = \dfrac{-1 + \sqrt{5}}{2}$

which gives

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

A Level   ## Example Questions

$2\sin x - \tan x = 0$

is equivalent to

$2\sin x - \dfrac{\sin x}{\cos x} = 0$

or

$2\sin x \cos x - \sin x = 0$

which can be factorised to

$(2\cos x - 1)\sin x = 0$

so

$\sin x = 0$ or $\cos x = \dfrac{1}{2}$

$\sin x = 0$ has solutions at $x = 0, \pi, 2\pi$

and

$\cos x = \dfrac{1}{2}$ has solutions at $x = \dfrac{\pi}{3}, \dfrac{5\pi}{3}$

So, there are five solutions, $x = 0, \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3}, 2\pi$.

$(\sin x + \cos x)(\sin x - \cos x)$

$= \sin ^2 x - \cos ^2 x$

$= 2\sin ^2 x - 1$

so we require

$\sin x = ±\dfrac{1}{\sqrt{2}}$

which gives us the set of solutions

$x = 45°, 135°, 225°, 315°$

Use the first identity, $\tan x \equiv \dfrac{\sin x}{\cos x}$

For each value of $x = 90°, 270°, 450°, ..., 90° ± 180°n$, we have $\cos x = 0$.

Since we cannot divide by $0$, we must have $\tan x$ undefined at these values.

A Level

A Level

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