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New Zealand
Level 8 - NCEA Level 3

Prove and apply other trigonometric identities


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(xy) $=$= $\sin x\cos y-\cos x\sin y$sinxcosycosxsiny
$\cos\left(x+y\right)$cos(x+y) $=$= $\cos x\cos y-\sin x\sin y$cosxcosysinxsiny
$\cos\left(x-y\right)$cos(xy) $=$= $\cos x\cos y+\sin x\sin y$cosxcosy+sinxsiny
$\tan\left(x+y\right)$tan(x+y) $=$= $\frac{\tan x+\tan y}{1-\tan x\tan y}$tanx+tany1tanxtany
$\tan\left(x-y\right)$tan(xy) $=$= $\frac{\tan x-\tan y}{1+\tan x\tan y}$tanxtany1+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)}$2tanx1tan2(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.



Manipulate trigonometric expressions


Apply trigonometric methods in solving problems

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