Trigonometric Identities

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Simplify expressions with reciprocals using complementary results

Lesson

In a right-angled triangle, it is clear that $\sin\theta=\cos\left(90^\circ-\theta\right)$`s``i``n``θ`=`c``o``s`(90°−`θ`) and similarly $\cos\theta=\sin\left(90^\circ-\theta\right)$`c``o``s``θ`=`s``i``n`(90°−`θ`). This was mentioned in an earlier chapter.

Also, it can be established from the definitions that

$\cot\theta=\tan\left(90^\circ-\theta\right)$`c``o``t``θ`=`t``a``n`(90°−`θ`) and $\tan\theta=\cot\left(90^\circ-\theta\right),$`t``a``n``θ`=`c``o``t`(90°−`θ`),

$\csc\theta=\sec\left(90^\circ-\theta\right)$`c``s``c``θ`=`s``e``c`(90°−`θ`) and $\sec\theta=\csc\left(90^\circ-\theta\right)$`s``e``c``θ`=`c``s``c`(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 $\tan\left(\frac{\pi}{2}-\theta\right)$`t``a``n`(π2−`θ`) we cannot use the tangent angle difference identity directly because $\tan\frac{\pi}{2}$`t``a``n`π2 is undefined. But, we can revert to the definition of the tangent function and write

$\tan\left(\frac{\pi}{2}-\theta\right)=\frac{\sin\left(\frac{\pi}{2}-\theta\right)}{\cos\left(\frac{\pi}{2}-\theta\right)}=\frac{\sin\frac{\pi}{2}\cos\theta-\cos\frac{\pi}{2}\sin\theta}{\cos\frac{\pi}{2}\cos\theta+\sin\frac{\pi}{2}\sin\theta}$`t``a``n`(π2−`θ`)=`s``i``n`(π2−`θ`)`c``o``s`(π2−`θ`)=`s``i``n`π2`c``o``s``θ`−`c``o``s`π2`s``i``n``θ``c``o``s`π2`c``o``s``θ`+`s``i``n`π2`s``i``n``θ`

Now, we can use the facts that $\sin\frac{\pi}{2}=1$`s``i``n`π2=1 and $\cos\frac{\pi}{2}=0$`c``o``s`π2=0 to simplify this expression and conclude that

$\tan\left(\frac{\pi}{2}-\theta\right)=\frac{\cos\theta}{\sin\theta}=\cot\theta$`t``a``n`(π2−`θ`)=`c``o``s``θ``s``i``n``θ`=`c``o``t``θ`.

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

Why not see if you can determine them?

Rewrite $\sec35^\circ15'$`s``e``c`35°15′ in terms of its cofunction.

The required cofunction is cosecant. We have $\sec35^\circ15'=\csc\left(90^\circ-35^\circ15'\right)$`s``e``c`35°15′=`c``s``c`(90°−35°15′). Hence, $\sec35^\circ15'$`s``e``c`35°15′ is the same as $\csc54^\circ45'$`c``s``c`54°45′.

Rewrite $\sin0.67$`s``i``n`0.67 in terms of its cofunction with the argument correct to two decimal places.

The required cofunction is cosine. We have $\sin0.67=\cos\left(\frac{\pi}{2}-0.67\right)$`s``i``n`0.67=`c``o``s`(π2−0.67). Thus, $\sin0.67\approx\cos0.90$`s``i``n`0.67≈`c``o``s`0.90

Remember!

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

Express $\sin\left(\pi+\theta\right)-\sin\left(\pi-\theta\right)$`s``i``n`(π+`θ`)−`s``i``n`(π−`θ`) in simplest form.

Rewrite $\sin18^\circ23'$`s``i``n`18°23′ in terms of its cofunction.

Rewrite $\sec\frac{\pi}{9}$`s``e``c`π9 in terms of its cofunction.

Manipulate trigonometric expressions

Apply trigonometric methods in solving problems