Simplify expressions with reciprocals using complementary results

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(π2​−θ) 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(π2​−θ)=sin(π2​−θ)cos(π2​−θ)​=sinπ2​cosθ−cosπ2​sinθcosπ2​cosθ+sinπ2​sinθ​

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(π2​−θ)=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(π2​−0.67). Thus, $\sin0.67\approx\cos0.90$sin0.67≈cos0.90

 

More Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3