7.05 Trigonometric identities
We have been simplifying algebraic expressions for a long time now. If we were asked to simplify $\frac{x}{x}$xx, we would say the answer is $1$1 (so long as $x$x is not zero). So, what if we were asked to simplify $\frac{\sin x}{\sin x}$sinxsinx?
It doesn't matter what size the angle of $x$x will be, the numerator and denominator of the fraction will be the same, and so the fraction will equal $1$1, so long as the denominator is not zero. $\frac{\sin30^\circ}{\sin30^\circ}=1$sin30°sin30°=1, $\frac{\sin250^\circ}{\sin250^\circ}=1$sin250°sin250°=1 and so on. This is an identity in the sense that it is true for every value of $x$x that makes the fraction well-defined. And so, the expression $\frac{\sin x}{\sin x}$sinxsinx, just like $\frac{x}{x}$xx, is always equal to $1$1.
Hence, an identity is defined as an equation/relationship that is true for every value of the variable that we may input.
Trigonometric identities
We have covered a range of important trigonometric identities in this topic so far, including identities for $\tan\theta$tanθ, $\cot\theta$cotθ, the reciprocals and the complementary relationships. These identities allow us to manipulate terms in expressions into different forms which enables us to undertake further mathematical operations which would be otherwise impossible.
$\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ
$\cot\theta=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}$cotθ=1tanθ=cosθsinθ
$\sec\theta=\frac{1}{\cos\theta}$secθ=1cosθ
$\csc\theta=\frac{1}{\sin\theta}$cscθ=1sinθ
$\cos\theta=\sin\left(90^\circ-\theta\right)$cosθ=sin(90°−θ) and $\sin\theta=\cos\left(90^\circ-\theta\right)$sinθ=cos(90°−θ)
$\cot\theta=\tan\left(90^\circ-\theta\right)$cotθ=tan(90°−θ) and $\tan\theta=\cot\left(90^\circ-\theta\right)$tanθ=cot(90°−θ)
$\csc\theta=\sec\left(90^\circ-\theta\right)$cscθ=sec(90°−θ) and $\sec\theta=\csc\left(90^\circ-\theta\right)$secθ=csc(90°−θ)
We can simplify trigonometric expressions using any of the four operations, by removing common factors or by using the identities above to substitute in equivalent expressions. Let's have a look at an example.
Worked example
Example 1
Find a simpler way of writing $\frac{\sin^2A+\sin A\cos A}{\cos^2A+\sin A\cos A}$sin2A+sinAcosAcos2A+sinAcosA.
Think: We can take a common factor out of the numerator and out of the denominator, so let's start there.
Do: The numerator has a common factor of $\sin A$sinA and the denominator has a common factor of $\cos A$cosA, so we can write the expression as:
$\frac{\sin A\left(\sin A+\cos A\right)}{\cos A\left(\cos A+\sin A\right)}$sinA(sinA+cosA)cosA(cosA+sinA)
On canceling the common term from the numerator and the denominator, we have:
$\frac{\sin A}{\cos A}=\tan A$sinAcosA=tanA
Practice questions
Question 1
Simplify $\csc\theta\tan\theta\cos\theta$cscθtanθcosθ.
Question 2
Simplify $\sin\left(90^\circ-y\right)\times\tan y$sin(90°−y)×tany.
Pythagorean identities
Pythagoras' theorem relates the three sides of a right-angled triangle. We can use expressions for the sides using trigonometric ratios to create three important identities.
Below is a right-angled triangle with two sides $a$a and $b$b, and hypotenuse $c$c. Let the angle opposite side $a$a be $\alpha$α.
We know that $a^2+b^2=c^2$a2+b2=c2 by Pythagoras' theorem. If we now divide the equation through by $c^2$c2, we obtain the form $\frac{a^2}{c^2}+\frac{b^2}{c^2}=1$a2c2+b2c2=1.
The fractions $\frac{a}{c}$ac and $\frac{b}{c}$bc are recognised as $\sin\alpha$sinα and $\cos\alpha$cosα respectively. Therefore, we can replace the fractions by the trigonometric ratios to obtain:
$\sin^2\alpha+\cos^2\alpha\equiv1$sin2α+cos2α≡1
The $\equiv$≡ sign can be used because this is an identity, meaning it is true whatever the value of $\alpha$α. You would have seen this symbol when writing proofs for congruent triangles, which were triangles that were identical in every way. In practice, it is fine for you to use a regular "equals" sign when writing solutions in this topic, but it is good to know how, and why, you might use the equivalence symbol.
To be convinced that the identity holds for angles of any magnitude, we can inspect the unit circle diagram and apply Pythagoras's theorem to the triangles that are formed. Note that although the values of $\sin\alpha$sinα and $\cos\alpha$cosα may be negative, their squares are not.
Now that we've obtained the fact that $\sin^2\alpha+\cos^2\alpha\equiv1$sin2α+cos2α≡1 we can divide both sides by $\sin^2\alpha$sin2α to obtain the following identity in terms of $\cot\alpha$cotα and $\csc\alpha$cscα:
| $\frac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha}$sin2α+cos2αsin2α | $\equiv$≡ | $\frac{1}{\sin^2\alpha}$1sin2α | (Dividing both sides by $\sin^2\alpha$sin2α) |
| $\frac{\sin^2\alpha}{\sin^2\alpha}+\frac{\cos^2\alpha}{\sin^2\alpha}$sin2αsin2α+cos2αsin2α | $\equiv$≡ | $\frac{1}{\sin^2\alpha}$1sin2α | (Breaking up the fraction) |
| $1+\cot^2\alpha$1+cot2α | $\equiv$≡ | $\csc^2\alpha$csc2α | (Simplifying each fraction) |
If we repeat the above process, but now instead divide both sides by $\cos^2\alpha$cos2α, then we obtain the following identity in terms of $\tan\alpha$tanα and $\sec\alpha$secα:
| $\frac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}$sin2α+cos2αcos2α | $\equiv$≡ | $\frac{1}{\cos^2\alpha}$1cos2α | (Dividing both sides by $\cos^2\alpha$cos2α) |
| $\frac{\sin^2\alpha}{\cos^2\alpha}+\frac{\cos^2\alpha}{\cos^2\alpha}$sin2αcos2α+cos2αcos2α | $\equiv$≡ | $\frac{1}{\cos^2\alpha}$1cos2α | (Breaking up the fraction) |
| $\tan^2\alpha+1$tan2α+1 | $\equiv$≡ | $\sec^2\alpha$sec2α | (Simplifying each fraction) |
These three identities are stated again below. The first is given to you on your examination reference sheet, but the two that follow from it are not. It is important that you commit these to memory or are able to derive them!
$\sin^2\theta+\cos^2\theta\equiv1$sin2θ+cos2θ≡1
$\tan^2\theta+1\equiv\sec^2\theta$tan2θ+1≡sec2θ
$1+\cot^2\theta\equiv\csc^2\theta$1+cot2θ≡csc2θ
It is expected that you can rearrange these identities to make any term in an expression the subject so that you can then substitute it into another expression. In Example 3 below, we can also see a very clever way to eliminate trigonometric ratios from our answer altogether using the Pythagorean identities.
Worked examples
Example 2
Simplify the expression $\left(\cos\beta+1\right)\left(\cos\beta-1\right)$(cosβ+1)(cosβ−1) and write it in terms of the sine function.
Think: Notice the expression is the factorised version of the difference of two squares. So let's expand, giving us squared terms, which might require the use of Pythagorean identities to further simplify.
Do: Expanding the brackets gives:
| $\left(\cos\beta+1\right)\left(\cos\beta-1\right)$(cosβ+1)(cosβ−1) | $=$= | $\cos^2\beta-1$cos2β−1 |
We can find an equivalent expression for this by rearranging $\sin^2\beta+\cos^2\beta=1$sin2β+cos2β=1. When we do this, we find that:
| $-\left(1-\cos^2\beta\right)$−(1−cos2β) | $=$= | $-\sin^2\beta$−sin2β |
Example 3
Given that $x=4\sin\theta$x=4sinθ and $y=5\cos\theta$y=5cosθ, find a relation between $x$x and $y$y that does not involve the trigonometric functions.
Think: In order to use the Pythagorean identity, we need the squares of the trigonometric functions.
Do: So let's start squaring the expressions for $x$x and $y$y. Squaring $x$x gives us:
| $x^2$x2 | $=$= | $16\sin^2\theta$16sin2θ |
And:
| $y^2$y2 | $=$= | $25\cos^2\theta$25cos2θ |
Then:
| $\frac{x^2}{16}$x216 | $=$= | $\sin^2\theta$sin2θ |
And:
| $\frac{y^2}{25}$y225 | $=$= | $\cos^2\theta$cos2θ |
We add these two equations to obtain:
| $\frac{x^2}{16}+\frac{y^2}{25}$x216+y225 | $=$= | $\sin^2\theta+\cos^2\theta$sin2θ+cos2θ |
And finally:
| $\frac{x^2}{16}+\frac{y^2}{25}$x216+y225 | $=$= | $1$1 |
Practice questions
Question 3
Simplify $\frac{1-\cos^2\left(\theta\right)}{1-\sin^2\left(\theta\right)}$1−cos2(θ)1−sin2(θ).
Question 4
Simplify $\frac{1}{1+\sec\theta}-\frac{1}{1-\sec\theta}$11+secθ−11−secθ.
Setting out a trigonometric proof
It is important to remember that trigonometric identities aren't equations to solve. For this reason, we might be asked to verify that an identity is correct, and we do this by working with the left and right-hand sides of the identities separately. It doesn't matter which side you start with, but we usually choose the side that looks more complicated (or has more terms) and we see if we can manipulate it until it is equivalent to the other side. Use the notation "LHS =" or "RHS = " in your solutions to be clear where you have chosen to start.
Below is an example of the setting out you should try to use in your proofs. It also shows us a useful trick when you can't seem to find a way to proceed, which is to multiply by a convenient (clever!) representation of $1$1.
Worked example
Example 4
Prove that $\frac{1}{\sin x+\cos x}=\frac{\sin x+\cos x}{1+2\sin x\cos x}$1sinx+cosx=sinx+cosx1+2sinxcosx.
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) | |
| $=$= | $\frac{\sin x+\cos x}{1+2\sin x\cos x}$sinx+cosx1+2sinxcosx | (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.
Practice questions
Question 5
By starting with the left hand side, prove that $\frac{1-\cos^2\left(\theta\right)}{\sin\theta}=\sin\theta$1−cos2(θ)sinθ=sinθ.
Question 6
Prove that $\cot\theta\left(\sec^2\left(\theta\right)-1\right)=\tan\theta$cotθ(sec2(θ)−1)=tanθ.