Lesson

Imagine a right-angled triangle with one of the acute angles labelled $\phi$`ϕ`. If the side opposite the angle $\phi$`ϕ` has length $a$`a` and the side adjacent to the angle $\phi$`ϕ` has length $b$`b`, then we write $\tan\phi=\frac{a}{b}$`t``a``n``ϕ`=`a``b`. But this is the same as the equivalent fraction $\frac{\frac{a}{c}}{\frac{b}{c}}$`a``c``b``c` obtained by dividing the numerator and the denominator by the number $c$`c`.

If we now let $c$`c` be the hypotenuse of the triangle, we see that the complicated looking fraction is just $\frac{\sin\phi}{\cos\phi}$`s``i``n``ϕ``c``o``s``ϕ`. Hence, we can write the identity

$\tan\phi\equiv\frac{\sin\phi}{\cos\phi}$`t``a``n``ϕ`≡`s``i``n``ϕ``c``o``s``ϕ`

(This is an identity in the sense that it is true for every value of $\phi$`ϕ`.)

Another very useful identity is obtained in a similar way. In the same triangle, we have the statement of Pythagoras's theorem $a^2+b^2\equiv c^2$`a`2+`b`2≡`c`2, which remains true for a given $c$`c` whatever the values of $a$`a` and $b$`b`, provided the triangle continues to be right-angled.

On dividing both sides of this equivalence by $c^2$`c`2, we obtain $\frac{a^2}{c^2}+\frac{b^2}{c^2}\equiv1$`a`2`c`2+`b`2`c`2≡1. We recognise this as just

$\sin^2\phi+\cos^2\phi\equiv1$`s``i``n`2`ϕ`+`c``o``s`2`ϕ`≡1.

This can now be divided through by $\cos^2\phi$`c``o``s`2`ϕ` to obtain a further identity involving the tangent function:

$\tan^2\phi+1\equiv\frac{1}{\cos^2\phi}=\sec^2\phi$`t``a``n`2`ϕ`+1≡1`c``o``s`2`ϕ`=`s``e``c`2`ϕ`

Find a simpler way of writing $\frac{\sin^2A+\sin A\cos A}{\cos^2A+\sin A\cos A}$`s``i``n`2`A`+`s``i``n``A``c``o``s``A``c``o``s`2`A`+`s``i``n``A``c``o``s``A`.

Taking a common factor out of the numerator and out of the denominator, we can write the expression as

$\frac{\sin A\left(\sin A+\cos A\right)}{\cos A\left(\cos A+\sin A\right)}$`s``i``n``A`(`s``i``n``A`+`c``o``s``A`)`c``o``s``A`(`c``o``s``A`+`s``i``n``A`).

On cancelling the common term from the numerator and the denominator, we have $\frac{\sin A}{\cos A}=\tan A$`s``i``n``A``c``o``s``A`=`t``a``n``A`.

Often, students are asked to verify an identity. This is usually done by rearranging or simplifying one side of the identity until it looks identical to the other side. The working above essentially proves the identity

$\frac{\sin^2A+\sin A\cos A}{\cos^2A+\sin A\cos A}\equiv\tan A$`s``i``n`2`A`+`s``i``n``A``c``o``s``A``c``o``s`2`A`+`s``i``n``A``c``o``s``A`≡`t``a``n``A`.

If $5\sin x=13\cos x$5`s``i``n``x`=13`c``o``s``x` and $x$`x` is between $0$0 and $2\pi$2π, what are the possible values of $x$`x`?

From $5\sin x=13\cos x$5`s``i``n``x`=13`c``o``s``x` we obtain $\frac{\sin x}{\cos x}=\frac{13}{5}=2.6$`s``i``n``x``c``o``s``x`=135=2.6

That is, $\tan x=2.6$`t``a``n``x`=2.6. Therefore, in the first quadrant, $x=\arctan2.6\approx1.2036$`x`=`a``r``c``t``a``n`2.6≈1.2036 or $\frac{\pi}{2.6101}$π2.6101. There should also be a third-quadrant solution since $\tan x=\tan\left(\pi+x\right)$`t``a``n``x`=`t``a``n`(π+`x`). That is, the second solution is$\pi+\frac{\pi}{2.6101}=\frac{3.6101\times\pi}{2.6101}$π+π2.6101=3.6101×π2.6101.

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

BYes

ANo

B

.

Given that $\sin x=\frac{4}{5}$`s``i``n``x`=45 and $\cos x=\frac{3}{5}$`c``o``s``x`=35, find $\tan x$`t``a``n``x`.

Find the value of $x$`x` when

$8\sin x=15\cos x$8`s``i``n``x`=15`c``o``s``x`.

Round your answer to the nearest degree.

Apply trigonometric relationships, including the sine and cosine rules, in two and three dimensions