New Zealand
Level 8 - NCEA Level 3

# Simplify expressions with reciprocals using complementary results

Lesson

In a right-angled triangle, it is clear that $\sin\theta=\cos\left(90^\circ-\theta\right)$sinθ=cos(90°θ) and similarly $\cos\theta=\sin\left(90^\circ-\theta\right)$cosθ=sin(90°θ). This was mentioned in an earlier chapter.

Also, it can be established from the definitions that

$\cot\theta=\tan\left(90^\circ-\theta\right)$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)

The trigonometric functions paired in this way are called cofunctions.

The same relations hold for angles of any magnitude. For example, to find the way to re-write $\tan\left(\frac{\pi}{2}-\theta\right)$tan(π2θ) we cannot use the tangent angle difference identity directly because $\tan\frac{\pi}{2}$tanπ2 is undefined. But, we can revert to the definition of the tangent function and write

$\tan\left(\frac{\pi}{2}-\theta\right)=\frac{\sin\left(\frac{\pi}{2}-\theta\right)}{\cos\left(\frac{\pi}{2}-\theta\right)}=\frac{\sin\frac{\pi}{2}\cos\theta-\cos\frac{\pi}{2}\sin\theta}{\cos\frac{\pi}{2}\cos\theta+\sin\frac{\pi}{2}\sin\theta}$tan(π2θ)=sin(π2θ)cos(π2θ)=sinπ2cosθcosπ2sinθcosπ2cosθ+sinπ2sinθ

Now, we can use the facts that $\sin\frac{\pi}{2}=1$sinπ2=1 and $\cos\frac{\pi}{2}=0$cosπ2=0 to simplify this expression and conclude that

$\tan\left(\frac{\pi}{2}-\theta\right)=\frac{\cos\theta}{\sin\theta}=\cot\theta$tan(π2θ)=cosθsinθ=cotθ.

The other cofunction identities can be confirmed in a similar way.

Why not see if you can determine them?

#### Example 1

Rewrite $\sec35^\circ15'$sec35°15 in terms of its cofunction.

The required cofunction is cosecant. We have $\sec35^\circ15'=\csc\left(90^\circ-35^\circ15'\right)$sec35°15=csc(90°35°15). Hence, $\sec35^\circ15'$sec35°15 is the same as $\csc54^\circ45'$csc54°45.

#### Example 2

Rewrite $\sin0.67$sin0.67 in terms of its cofunction with the argument correct to two decimal places.

The required cofunction is cosine. We have $\sin0.67=\cos\left(\frac{\pi}{2}-0.67\right)$sin0.67=cos(π20.67). Thus, $\sin0.67\approx\cos0.90$sin0.67cos0.90

Remember!

Note that in the absence of a $^\circ$° sign, we always assume the angle is given in radians.

#### More Worked Examples

##### QUESTION 1

Express $\sin\left(\pi+\theta\right)-\sin\left(\pi-\theta\right)$sin(π+θ)sin(πθ) in simplest form.

##### QUESTION 2

Rewrite $\sin18^\circ23'$sin18°23 in terms of its cofunction.

##### QUESTION 3

Rewrite $\sec\frac{\pi}{9}$secπ9 in terms of its cofunction.

### Outcomes

#### M8-6

Manipulate trigonometric expressions

#### 91575

Apply trigonometric methods in solving problems