Trigonometric Identities

Q: What are the three trigonometric identities?

sinx/cosx=tanx, sin^2(x)+cos^2(x)=1. 1/cosx=secx

Q: How do you solve trigonometric identities easily?

Simply rearrange to the identities listed above and substitute them back in.

Q: How do we verify trigonometric identities?

Drawing a diagram reveals why each identity works. Regular SOHCAHTOA can show what’s going on.

Q: What are trigonometric identities used for?

They can help us solve larger trigonometric equations that cannot be solved otherwise.

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Trigonometric identities are important to work through a variety of problems and advanced Equations. They allow us to simplify many problems and make situations easier.

What is the main set of trigonometric identities?

There are two main formulaic identities that must be learnt to prove and solve other Equations. These are:

and $\frac{\sin^{2} x}{\cos^{2} x} = \tan x$

Let’s prove these identities starting with $\sin^{2} x + \cos^{2} x = 1$ .

Proof:

Firstly let’s draw a triangle with angle θ.

Trigonometric identities Triangle StudySmarter

General Triangle of angle θ

Now if we write out expressions for a and b using SOHCAHTOA we get:

a = c \sin θ b = c \cos θ

Therefore:

\frac{a}{c} = \sin θ \frac{b}{c} = \cos θ

Now if we square both of these expressions for sin and cos we get:

\frac{a^{2}}{c^{2}} = \sin^{2} θ \frac{b^{2}}{c^{2}} = \cos^{2} θ

Summing these we get:

\sin^{2} θ + \cos^{2} θ = \frac{a^{2} + b^{2}}{c^{2}}

By Pythagoras' theorem:

a^{2} + b^{2} = c^{2}

Therefore:

\frac{a^{2} + b^{2}}{c^{2}} = \frac{c^{2}}{c^{2}} = 1 \sin^{2} θ + \cos^{2} θ = 1

Now let’s move on to proving $\frac{\sin x}{\cos x} = \tan x$ . The first half of this Proof is identical to the Proof above.

PROOF:

Firstly let’s draw a triangle with angle θ.

Now if we write out expressions for a and b using SOHCAHTOA we get:

Now if we divide these two expressions for sin and cos:

\frac{\sin θ}{\cos θ} = \frac{(\frac{a}{c})}{(\frac{b}{c})}

= \frac{a}{c} \times \frac{c}{b} = \frac{a}{b}

This is an expression for the opposite side over the adjacent side, therefore:

\frac{a}{b} = \tan θ

Therefore:

\frac{\sin θ}{\cos θ} = \tan θ

Now let’s look at some worked examples where trigonometric identities can be applied.

Worked examples using trigonometric identities

Solve the equation $4 \sin^{2} x + 8 \cos x - 7 = 0$ for $0 \leq x \leq 180 .$

SOLUTION:The first thing to do would be to substitute

1 - \cos^{2} x

for

\sin^{2} x

.The equation now ends up being

4 (1 - \cos^{2} x) + 8 \cos x - 7 = 0

.Simplifying this further:

4 - 4 \cos^{2} x + 8 \cos x - 7 = 0

4 \cos^{2} x - 8 \cos x + 3 = 0

Now we can solve this like a quadratic by taking

y = \cos x

4 y^{2} - 8 y + 3 = 0 (2 y - 1) (2 y - 3) = 0 y = 0.5 o r y = 1.5

Now we need to do x = cos^-1(y)We can only perform cos^-1(0.5)=60°This is because 1.5 > 1 so we cannot perform a cos^-1 function of this.So the only answer is 60°.

Let's look at another example of rearranging trigonometric identities.

Show that the equation $2 \sin x = \frac{(4 \cos x - 1)}{\tan x}$ can be written as $6 \cos^{2} x - \cos x - 2 = 0 .$

SOLUTION:Firstly let’s rearrange to get rid of any denominators.

2 \sin x \tan x = 4 \cos x - 1

Now let’s replace

\tan x

with

\frac{\sin x}{\cos x}

2 \sin x \frac{\sin x}{\cos x} = 4 \cos x - 1

\frac{2 \sin^{2} x}{\cos x} = 4 \cos x - 1

Now get rid of the denominator by multiplying through by

\cos x

2 \sin^{2} x = 4 \cos^{2} x - \cos x

Now replace

\sin^{2} x

with

1 - \cos^{2} x

2 (1 - \cos^{2} x) = 4 \cos^{2} x - \cos x 2 - 2 \cos^{2} x = 4 \cos^{2} x - \cos x

Now rearrange this equation:

2 = 6 \cos^{2} x - \cos x 6 \cos^{2} x - \cos x - 2 = 0

QED

What other trigonometric identities can we derive?

Firstly we need to know three new bits of terminology:

$s e c x = \frac{1}{\cos x} \cos e c x = \frac{1}{\sin x} c o t x = \frac{1}{\tan x}$

These are all reciprocals of standard sin, cos and tan.

Deriving new identities

Now let’s look at the identity $\sin^{2} x + \cos^{2} x = 1$ :

If we divide the entire equation by

\cos^{2} (x)

we get:

\frac{\sin^{2} x}{\cos^{2} x} + \frac{\cos^{2} x}{\cos^{2} x} = \frac{1}{\cos^{2} x}

Now using the identity

\frac{\sin x}{\cos x} = \tan x

\tan^{2} x + 1 = s e c^{2} x

This is our first new identity. Now if we divide our entire equation by

\sin^{2} x

\frac{\sin^{2} x}{\sin^{2} x} + \frac{\cos^{2} x}{\sin^{2} x} = \frac{1}{\sin^{2} x}

Now using the identity

\frac{\sin x}{\cos x} = \tan x

, so :

1 + \frac{1}{\tan^{2} x} = \frac{1}{\sin^{2} x} 1 + c o t^{2} x = \cos e c^{2} x

Now we have our two new identities:

\tan^{2} x + 1 = s e c^{2} x c o t^{2} x + 1 = \cos e c^{2} x

Let’s see them in action in some worked examples.

Worked examples of new identities

Solve, for 0 ≤ θ < 360°, the equation:

2 \tan^{2} x + s e c x = 1

to 1 dp.

Graph of y=cosx. Image: Ruben Verhaegh, CC BY-SA 4.0

We can see that if we perform the identity $\cos x = \cos (360 - x)$ , the other value of $x$ is $360 - 131.8 = 228.2$ .

Then we need to perform $\cos^{- 1} (1) = 0$ , again using the identity $\cos x = \cos (360 - x)$ , $x = 360$ .

So to 1 decimal place our 4 solutions in degrees are:

$x = 131.8, x = 228.2, x = 0, x = 360$