# Basic Trig Identities

## Basic Trig Identities Revision

**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.

**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.

**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.

**Example: Adapting Equations**

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

## Basic Trig Identities Example Questions

**Question** **1:** Find the solutions for 2\sin x - \tan x = 0 in the interval 0 \leq x \leq 2\pi.

**[4 marks]**

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.

**Question 2:** Show that (\sin x + \cos x)(\sin x - \cos x) = 0 has solutions x = 45°, 135°, 225°, 315° for the interval 0° \leq x \leq 360°.

**[3 marks]**

(\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°

**Question 3:** Explain briefly why \tan x has no value when x = 90°, 270°, 450°, ..., 90° ± 180°n.

**[3 marks]**

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.

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