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CanadaON
Grade 12

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(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.

Outcomes

12F.B.3.3

Recognize that trigonometric identities are equations that are true for every value in the domain, prove trigonometric identities through the application of reasoning skills, using a variety of relationships and verify identities using technology

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