Trigonometric Function

Radians

As in the figure, we let the origin O. Take points ${\rm A}(1,0), {\rm P}(x,y)$ on the unit circle $x^{2} + y^{2} = 1$. Now set $\angle {\rm POA}$ as $\theta$. The angle $\angle {\rm POA}$ for which the arclength of AP is 1 is called 1 radian and denoted by 1rad^1.3.

NOTE Suppose that $\angle {\rm POA}$ is $\theta$ degree and $x$ radian. Then

$$\frac{\theta}{360} = \frac{x}{2\pi}$$

Also, basic angles are measured in degrees and radians to give

Degree	0	30	45	60	90	120	150	180	360
Radian	0	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\frac{2\pi}{3}$	$\frac{5\pi}{6}$	$\pi$	$2\pi$

Trigonometric Functions

Suppose that $\theta = \angle {\rm POA}$. Then the following functions of $\theta$ are called Trigonometric functions.

$$\frac{y}{r} = \sin{\theta}\hskip 0.3cm \frac{x}{r} = \cos{\theta} \hskip 0.3cm \frac{y}{x} = \tan{\theta}$$

NOTE As $\theta$ chages the value, the point P$(x,y)$ and the shape of the right triangle $\triangle$ OPH changes.

Example 1..5   Find the value of the followings.
1. $\sin{\frac{\pi}{3}}$ 2. $\sin{(-\frac{\pi}{3})}$ 3. $\cos{\frac{3\pi}{4}}$ 4. $\tan{(-\frac{3\pi}{4})}$

SOLUTION 1. Draw a unit circle with the origin O and draw a line OP with $\theta = \frac{\pi}{3}$. Then the value of $y$ coordinate of P is equal to $\sin{\frac{\pi}{3}}$.

$$\sin{\frac{\pi}{3}} = \frac{\sqrt{3}}{2}\ensuremath{\ \blacksquare}$$

2. Similarly, the value of $y$ coordinate of P where $\theta = -\frac{\pi}{3}$ is equal to $\sin{-\frac{\pi}{3}}$. Thus we have

$$\sin{(-\frac{\pi}{3})} = -\frac{\sqrt{3}}{2}\ensuremath{\ \blacksquare}$$

3. The value of $x$ coordinate of P where $\theta = \frac{3\pi}{4}$ is equal to $\cos{\frac{3\pi}{4}}$. Thus we have

$$\cos{\frac{3\pi}{4}} = -\frac{1}{\sqrt{2}}\ensuremath{\ \blacksquare}$$

4. Stretch the line OP with $\theta = -\frac{3\pi}{4}$ so that $x$ coordinate is -1. Then the ratio of the values of $y$ coordinate and $x$ coordinate is $\tan{-\frac{3\pi}{4}}$. Thus we have

$$\tan{(-\frac{3\pi}{4})} = 1\ensuremath{\ \blacksquare}$$

Exercise 1..5   Find the value of the following.
1. $${\cos{\frac{5\pi}{4}}}$$ 2. $${\sin(-\frac{5\pi}{6})}$$ 3. $${\tan(-\frac{\pi}{3})}$$

SOLUTION 1. Draw a unit circle with the origin O and draw a line OP with $\theta = \frac{5\pi}{4}$. Then the value of $x$ coordinate of P is equal to $\cos{\frac{5\pi}{4}}$. Thus we have

$$\cos{\frac{5\pi}{4}} = -\frac{\sqrt{2}}{2}\ensuremath{\ \blacksquare}$$

2. Draw a unit circle with the origin O and draw a line OP with $\theta = -\frac{5\pi}{6}$. Then the value of $y$ coordinate of P is equal to $\sin(-\frac{5\pi}{6})$. Thus we have

$$\sin(-\frac{5\pi}{6}) = -\frac{1}{2}\ensuremath{\ \blacksquare}$$

3. Stretch the line OP with $\theta = -\frac{\pi}{3}$ so that $x$ coordinate is 1. Then the ratio of the values of $y$ coordinate and $x$ coordinate is $\tan{-\frac{\pi}{3}}$. Thus we have

$$\tan{(-\frac{\pi}{3})} = -\sqrt{3}\ensuremath{\ \blacksquare}$$

Basic Trig Identities

For all $\alpha,\beta$,
1. $${\cos^2{\alpha} + \sin^2{\alpha} = 1}$$
2. $${\sin(\alpha \pm \beta) = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}}$$
3. $${\cos(\alpha \pm \beta) = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}}$$

NOTE 1. Consider the point P$(x,y)$ on the unit circle. Then $\sqrt{x^2 + y^2} = \sqrt{\cos^{2}{\alpha} + \sin^{2}{\alpha}} = 1.$

2. ,3. Look at the figure, you will see $\sin(\alpha + \beta) = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}$

2. ,3. Write $\sin(\alpha - \cos{\beta})$ as $\sin(\alpha + (-\beta))$ and note that $\cos(-\beta) = \cos{\beta}, \sin(-\beta) = -\sin{\beta}$.

Example 1..6   Find the value of $\cos\frac{5\pi}{4}$.

SOLUTION

$$\cos{\frac{5\pi}{4}} = \cos(\pi + \frac{\pi}{4}) = \cos \pi \cos\frac{\pi}{4} - \sin{\pi}\sin{\frac{\pi}{4}}$$

$$= (-1)\cdot \frac{\sqrt{2}}{2} - 0\cdot \frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}} \ensuremath{\ \blacksquare}$$

Exercise 1..6   Find the value of $\sin{\frac{5\pi}{6}}$.

SOLUTION

$$\sin{\frac{5\pi}{6}} = \sin(\pi -\frac{\pi}{6}) = \sin\pi \cos\fr... ...pi}{6} - \cos \pi \sin{\frac{\pi}{6}} = \frac{1}{2} \ensuremath{\ \blacksquare}$$

Example 1..7   Find the value of $${\sin(\frac{7\pi}{12})}$$.

SOLUTION $\frac{7\pi}{12} = \frac{3\pi + 4\pi}{12} = \frac{\pi}{4} + \frac{\pi}{3}$. Now using trigonometric addition formula, we have

$$\sin(\frac{7\pi}{12}) = \sin(\frac{\pi}{4}+\frac{\pi}{3}) = \sin{\frac{\pi}{4}}\cos{\frac{\pi}{3}} + \cos{\frac{\pi}{4}}\sin{\frac{\pi}{3}}$$

$$= \frac{\sqrt{2}}{2}\cdot \frac{1}{2} + \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2} + \sqrt{6}}{4} \ensuremath{\ \blacksquare}$$

Exercise 1..7   Find the value of $${\cos(\frac{\pi}{8})}$$.

SOLUTION $\cos(\frac{\pi}{8} + \frac{\pi}{8}) = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. Using trigonometric addition formula

$$\cos(\frac{\pi}{8} + \frac{\pi}{8}) = \cos^{2}\frac{\pi}{8} - \si... ...os^{2}\frac{\pi}{8} - (1 - \cos^{2}\frac{\pi}{8}) = 2\cos^{2}\frac{\pi}{8} - 1.$$

Thus, $2\cos^{2}\frac{\pi}{8} = 1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2} }{2}$. $\cos^{2}\frac{\pi}{8} = \frac{2 + \sqrt{2}}{4}.$ Here, $\cos\frac{\pi}{8} > 0$, we have $\cos\frac{\pi}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$ $\ \blacksquare$