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.
| $\sin^2\left(x\right)+\cos^2\left(x\right)$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.
| $\sin\left(x+y\right)$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:
| $\sin2x$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 |