Topic 16 Trigonometric functions

16.1 From right triangles

Figure 16.1: A right triangle

Theorem 16.1 (Pythagoras's Theorem) \(a^2+b^2=c^2\)

Proof (non-examinable): \[\begin{align*} &&(a+b)^2 &= \text{area of the big square} = 4\left(\frac12ab\right) + c^2\\ \implies&&\quad a^2+2ab+b^2 &= 2ab+c^2\\ \implies&&\quad a^2+b^2&=c^2 \end{align*}\]

Returning to the figure of a right triangle above:

\(AB\) is called the hypotenuse.
\(AC\) is the side opposite to \(\theta\)
\(BC\) is the side adjacent to \(\theta\)

The definitions of the values of the trigonometric functions for \(\theta\) between \(0\) and \(\frac{\pi}{2}\):

\(\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{AC}{AB}\) (sine of \(\theta\))
\(\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{BC}{AB}\) (cosine of \(\theta\))
\(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{AC}{BC}\) (tangent of \(\theta\))
\(\sec\theta = \dfrac{1}{\cos\theta} = \dfrac{AB}{BC}\) (secant of \(\theta\))
\(\mathrm{cosec}\,\theta = \dfrac{1}{\sin\theta} = \dfrac{AB}{AC}\) (cosecant of \(\theta\))
\(\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{BC}{AC}\) (cotangent of \(\theta\))

Theorem 16.2 (Pythagoras's identity) \(\sin^2\theta+\cos^2\theta=1\)

Proof: \(BC^2+AC^2=AB^2\) (by Pythagoras’s Theorem)
\(\implies\) \(\left(\dfrac{BC}{AB}\right)^2+\left(\dfrac{AC}{AB}\right)^2=1\) (dividing by \(AB^2\))
\(\implies\) \(\cos^2\theta+\sin^2\theta = 1\) (by the definitions above)

Example 16.1 Prove that \(\dfrac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta\).

Solution. \[\begin{align*} \mathrm{LHS} &= \frac{1+\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\cos^2\theta}{\sin^2\theta}}\\ &= \frac{\sin^2\theta\,(\cos^2\theta+\sin^2\theta)}{\cos^2\theta\,(\sin^2\theta+\cos^2\theta)}\\ &=\frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta = \mathrm{RHS} \end{align*}\]

Example 16.2 Prove that \(\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}} = \mathrm{cosec}\,\theta-\cot\theta\).

Solution. \[\begin{align*} \mathrm{RHS} &= \frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta}= \frac{1-\cos\theta}{\sin\theta}.\\ \mathrm{LHS} &= \sqrt{\frac{(1-\cos\theta)(1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}} \\ &= \frac{\sqrt{(1-\cos\theta)^2}}{\sqrt{1-\cos^2\theta}} = \frac{1-\cos\theta}{\sqrt{\sin^2\theta}}\\ &= \frac{1-\cos\theta}{\sin\theta} = \mathrm{LHS}. \end{align*}\]

A table of very commonly used values:

\(\theta\) :————— \(\sin\theta\)	\(0\) :—–: \(0\)	\(\frac{\pi}6\) :——————: \(\frac12\)	\(\frac{\pi}4\) :——————: \(\frac1{\sqrt2}\)	\(\frac{\pi}3\) :——————: \(\frac{\sqrt3}2\)	\(\frac{\pi}2\) :—————: \(1\)

\(\cos\theta\)	\(1\)	\(\frac{\sqrt3}2\)	\(\frac1{\sqrt2}\)	\(\frac12\)	\(0\)
\(\tan\theta\)	\(0\)	\(\frac1{\sqrt3}\)	\(1\)	\(\sqrt3\)	not defined

How to derive some of these?

\(\theta=\frac{\pi}4\) implies \(AC=BC\), so \(\sin\theta=\cos\theta\).
So from Pythagoras’ Identity “\(\sin^2\theta+\cos^2\theta=1\)”, we get
\(2\sin^2\theta=1\) \(\implies\) \(\sin\theta=1/\sqrt2\).
\(\theta=\frac{\pi}{6}\): by doubling the triangle we get an equilateral one, thus
\(AC=\frac12AB\). Hence \(\sin\theta=\frac{AC}{AB}=\frac12\), and also
\(\cos\theta=\sqrt{1-\sin^2\theta} = \sqrt{\frac34} = \frac{\sqrt{3}}2\).

16.2 As functions

Given an angle \(\theta\), let \(P=(x,y)\) be the intersection point of the unit circle with the half-line through the origin making an angle \(\theta\) with the positive half of the \(x\)-axis.

Then, by the definition of trig functions above: \[\begin{align*} x&= \cos\theta,\\ y&= \sin\theta. \end{align*}\]

We can take this as the definition of \(\sin\) and \(\cos\) (and consequently of \(\tan\), …) if \(\theta\) is arbitrary (e.g. negative, or larger than \(\frac\pi2\)).

With this, we get basic trigonometric identities: \[\begin{align*} \cos(-\theta) &=\cos\theta, & \sin(-\theta) &=-\sin\theta \\ \cos(\theta+2\pi) &=\cos\theta & \sin(\theta+2\pi) &=\sin\theta\\ \cos(\theta+\pi) &= -\cos\theta & \sin(\theta+\pi) &= -\sin\theta\\ \tan(\theta+\pi) &= \tan\theta & \tan(-\theta) &= -\tan\theta \end{align*}\]