Trigonometry

Hong Kong

Stage 1 - Stage 3

Lesson

There are connections between the trigonometric functions that can make simplifications possible.

In a right-angled triangle, the two acute angles together make a right-angle. We say the acute angles are *complementary *(to one another). If the two acute angles have measures $\alpha$`α` and $\beta$`β`, then $\alpha+\beta=90^\circ$`α`+`β`=90° and so, $\beta=90^\circ-\alpha$`β`=90°−`α`.

The diagram below illustrates the following relationships.

$\cos\alpha=\frac{b}{h}=\sin\beta=\sin\left(90^\circ-\alpha\right)$`c``o``s``α`=`b``h`=`s``i``n``β`=`s``i``n`(90°−`α`)

$\sin\alpha=\frac{a}{h}=\cos\beta=\cos\left(90^\circ-\alpha\right)$`s``i``n``α`=`a``h`=`c``o``s``β`=`c``o``s`(90°−`α`)

$\cot\alpha=\frac{b}{a}=\tan\beta=\tan\left(90^\circ-\alpha\right)$`c``o``t``α`=`b``a`=`t``a``n``β`=`t``a``n`(90°−`α`)

Thus, the 'co' in complementary explains the meaning of *co*sine in relation to sine, and to *co*tangent in relation to tangent.

The statements

$\cos\alpha\equiv\sin\left(90^\circ-\alpha\right)$`c``o``s``α`≡`s``i``n`(90°−`α`),

$\sin\alpha\equiv\cos\left(90^\circ-\alpha\right)$`s``i``n``α`≡`c``o``s`(90°−`α`) and

$\cot\alpha\equiv\tan\left(90^\circ-\alpha\right)$`c``o``t``α`≡`t``a``n`(90°−`α`)

are called *identities* because they are true whatever the value of the angle $\alpha$`α`.

These identities are true not only in right-angled triangle trigonometry, but they also hold for angles of any size. This can be confirmed by thinking about the geometry in the unit circle diagram that is used for defining the trigonometric functions of angles of any magnitude.

Simplify the relation $\sin\left(90^\circ-\theta\right)=\sqrt{3}\sin\theta$`s``i``n`(90°−`θ`)=√3`s``i``n``θ`.

it will be a good plan to try to rearrange the equation so that the trigonometric functions are on one side and the coefficients are on the other. We divide both sides by $\sin\left(90^\circ-\theta\right)$`s``i``n`(90°−`θ`) and also by $\sqrt{3}$√3 to obtain $\frac{1}{\sqrt{3}}=\frac{\sin\theta}{\sin\left(90^\circ-\theta\right)}$1√3=`s``i``n``θ``s``i``n`(90°−`θ`). But, $\sin\left(90^\circ-\theta\right)$`s``i``n`(90°−`θ`) is just $\cos\theta$`c``o``s``θ`. So, the simplification we seek is

$\tan\theta=\frac{1}{\sqrt{3}}$`t``a``n``θ`=1√3.

We recognise an exact value for $\tan$`t``a``n` and conclude that $\theta=30^\circ$`θ`=30° if $\theta$`θ` is acute. You should check that there is also a third quadrant solution, $\theta=210^\circ$`θ`=210°.

By finding the ratio represented by $\sin\theta$`s``i``n``θ`, $\cos\theta$`c``o``s``θ` and $\tan\theta$`t``a``n``θ` in the given figure, we want to prove that $\frac{\sin\theta}{\cos\theta}=\tan\theta$`s``i``n``θ``c``o``s``θ`=`t``a``n``θ`.

Write down the expression for $\sin\theta$

`s``i``n``θ`.Write down the expression for $\cos\theta$

`c``o``s``θ`.Hence, form an expression for $\frac{\sin\theta}{\cos\theta}$

`s``i``n``θ``c``o``s``θ`.Write down the expression for $\tan\theta$

`t``a``n``θ`.Does $\frac{\sin\theta}{\cos\theta}=\tan\theta$

`s``i``n``θ``c``o``s``θ`=`t``a``n``θ`?Yes

ANo

B

Prove that $\frac{\tan x\cos x}{\sin x}=1$`t``a``n``x``c``o``s``x``s``i``n``x`=1.

Simplify the following expression using complementary angles:

$\frac{\sin51^\circ}{\cos39^\circ}$`s``i``n`51°`c``o``s`39°