The sum and difference identities can be used to generate further useful identities and they are often required when it is desired to simplify an expression or to write it in another form.
We could, for example, find a way of writing $\sin2x$sin2x without the double angle by noting that $2x$2x is just $x+x$x+x. Then,
| $\sin2x$sin2x | $=$= | $\sin\left(x+x\right)$sin(x+x) |
| $=$= | $\sin x\cos x+\cos x\sin x$sinxcosx+cosxsinx | |
| $=$= | $2\sin x\cos x$2sinxcosx |
Thus, we have derived the new identity
$\sin2x\equiv2\sin x\cos x$sin2x≡2sinxcosx
We can find an expression for $\cos2x$cos2x as follows:
$\cos2x=\cos\left(x+x\right)=\cos x\cos x-\sin x\sin x$cos2x=cos(x+x)=cosxcosx−sinxsinx. Therefore,
$\cos2x\equiv\cos^2x-\sin^2x$cos2x≡cos2x−sin2x.
Similarly, we find that
$\tan2x\equiv\frac{2\tan x}{1-\tan^2x}$tan2x≡2tanx1−tan2x.
We may wish to go further and find an expression for $\sin3x$sin3x without the triple angle. Since $\sin3x=\sin\left(x+2x\right)=\sin x\cos2x+\cos x\sin2x$sin3x=sin(x+2x)=sinxcos2x+cosxsin2x, and we alreadyhave the double angle formulas, we can write
$\sin3x=\sin x\left(\cos^2x-\sin^2x\right)+\cos x.2\sin x\cos x$sin3x=sinx(cos2x−sin2x)+cosx.2sinxcosx, or more simply,
$\sin3x\equiv\sin x\left(3\cos^2x-\sin^2x\right)$sin3x≡sinx(3cos2x−sin2x)
Exercises involving the angle sum and difference identities, and others, often ask for the manipulation of one side of a supposed identity in order to verify that the two sides are in fact the same.
Example
Verify that the statement is true.
$\tan\left(\theta+\alpha\right)\tan\left(\theta-\alpha\right)\equiv\frac{\tan^2\left(\theta\right)-\tan^2\left(\alpha\right)}{1-\tan^2\left(\theta\right)\tan^2\left(\alpha\right)}$tan(θ+α)tan(θ−α)≡tan2(θ)−tan2(α)1−tan2(θ)tan2(α)
We could manipulate the left and turn it into the right, or vice versa. Let's go from the right to the left.
The expression on the right can be factored. So,
| $\frac{\tan^2\left(\theta\right)-\tan^2\left(\alpha\right)}{1-\tan^2\left(\theta\right)\tan^2\left(\alpha\right)}$tan2(θ)−tan2(α)1−tan2(θ)tan2(α) | $=$= | $\frac{\left(\tan\theta+\tan\alpha\right)\left(\tan\theta-\tan\alpha\right)}{\left(1+\tan\theta\tan\alpha\right)\left(1-\tan\theta\tan\alpha\right)}$(tanθ+tanα)(tanθ−tanα)(1+tanθtanα)(1−tanθtanα) |
| $=$= | $\frac{\left(\tan\theta+\tan\alpha\right)}{\left(1-\tan\theta\tan\alpha\right)}.\frac{\left(\tan\theta-\tan\alpha\right)}{\left(1+\tan\theta\tan\alpha\right)}$(tanθ+tanα)(1−tanθtanα).(tanθ−tanα)(1+tanθtanα) | |
| $=$= | $\tan\left(\theta+\alpha\right).\tan\left(\theta-\alpha\right)$tan(θ+α).tan(θ−α) | |
| $=$= | Left Hand Side |
Worked Examples
QUESTION 1
By simplifying the left hand side of the identity , prove that $\tan\left(\theta+\alpha\right)\tan\left(\theta-\alpha\right)=\frac{\tan^2\left(\theta\right)-\tan^2\left(\alpha\right)}{1-\tan^2\left(\theta\right)\tan^2\left(\alpha\right)}$tan(θ+α)tan(θ−α)=tan2(θ)−tan2(α)1−tan2(θ)tan2(α).
QUESTION 2
If $\tan A+\tan B=-5$tanA+tanB=−5 and $\tan A\tan B=6$tanAtanB=6, prove that $\sin\left(A+B\right)=\cos\left(A+B\right)$sin(A+B)=cos(A+B).
QUESTION 3
Consider the equation $x^2-\sqrt{2}\sin\left(45^\circ+\theta\right)x+\sin\theta\cos\theta=0$x2−√2sin(45°+θ)x+sinθcosθ=0.
By first simplifying the coefficient of $x$x, solve for $x$x by factoring.