New Zealand
Level 8 - NCEA Level 3

# Prove and apply other trigonometric identities

Lesson

Trigonometric identities are useful as they allow us to draw connections between geometric contexts.

In addition to the definitions of $\tan x$tanx$\cot x$cotx$\sec x$secx and $\csc x$cscx, proofs in this set will also make use of the following identities.

Pythagorean identities
 $\sin^2\left(x\right)+\cos^2\left(x\right)$sin2(x)+cos2(x) $=$= $1$1 $\left(1\right)$(1) $\tan^2\left(x\right)+1$tan2(x)+1 $=$= $\sec^2\left(x\right)$sec2(x) $\left(2\right)$(2) $1+\cot^2\left(x\right)$1+cot2(x) $=$= $\operatorname{cosec}^2\left(x\right)$cosec2(x) $\left(3\right)$(3)

Dividing $\left(1\right)$(1) by $\cos^2\left(x\right)$cos2(x) or $\sin^2\left(x\right)$sin2(x) will give $\left(2\right)$(2) and $\left(3\right)$(3) respectively.

Compound angle formulae
 $\sin\left(x+y\right)$sin(x+y) $=$= $\sin x\cos y+\cos x\sin y$sinxcosy+cosxsiny $\sin\left(x-y\right)$sin(x−y) $=$= $\sin x\cos y-\cos x\sin y$sinxcosy−cosxsiny $\cos\left(x+y\right)$cos(x+y) $=$= $\cos x\cos y-\sin x\sin y$cosxcosy−sinxsiny $\cos\left(x-y\right)$cos(x−y) $=$= $\cos x\cos y+\sin x\sin y$cosxcosy+sinxsiny $\tan\left(x+y\right)$tan(x+y) $=$= $\frac{\tan x+\tan y}{1-\tan x\tan y}$tanx+tany1−tanxtany​ $\tan\left(x-y\right)$tan(x−y) $=$= $\frac{\tan x-\tan y}{1+\tan x\tan y}$tanx−tany1+tanxtany​

If $x=y$x=y in the above identities, then we obtain the following double angle formulae:

Double angle formulae
 $\sin2x$sin2x $=$= $2\sin x\cos x$2sinxcosx $\cos2x$cos2x $=$= $\cos^2\left(x\right)-\sin^2\left(x\right)$cos2(x)−sin2(x) $\tan2x$tan2x $=$= $\frac{2\tan x}{1-\tan^2\left(x\right)}$2tanx1−tan2(x)​

A common trick used in proofs is to multiply by a convenient representation of $1$1. In some cases, this trick will create an opportunity to use a known identity.

#### Worked example

Prove that $\frac{1}{\sin x+\cos x}=\frac{\sin x+\cos x}{1+\sin2x}$1sinx+cosx=sinx+cosx1+sin2x.

Think: There are currently no identities we can use to simplify this expression, however, multiplying by a convenient representation of $1$1 may present such an opportunity to us.

Do: Since our final answer has a $\sin x+\cos x$sinx+cosx in the numerator, let's start by multiplying by $\frac{\sin x+\cos x}{\sin x+\cos x}=1$sinx+cosxsinx+cosx=1.

 $LHS$LHS $=$= $\frac{1}{\sin x+\cos x}\times\frac{\sin x+\cos x}{\sin x+\cos x}$1sinx+cosx​×sinx+cosxsinx+cosx​ (Multiplying by $1$1) $=$= $\frac{\sin x+\cos x}{\left(\sin x+\cos x\right)^2}$sinx+cosx(sinx+cosx)2​ (Simplifying the multiplication) $=$= $\frac{\sin x+\cos x}{\sin^2\left(x\right)+\cos^2\left(x\right)+2\sin x\cos x}$sinx+cosxsin2(x)+cos2(x)+2sinxcosx​ (Expanding the denominator) $=$= $\frac{\sin x+\cos x}{1+\sin2x}$sinx+cosx1+sin2x​ (Applying identities) $=$= $RHS$RHS
Reflect: There is an infinite number of representations of $1$1 that we could have multiplied by, so it's important to let the desired $RHS$RHS inform us when we are trying to determine the best option.

### Outcomes

#### M8-6

Manipulate trigonometric expressions

#### 91575

Apply trigonometric methods in solving problems