Trigonometric Function

**Figure 1.12:** Radians
$\includegraphics[width=3.5cm]{SOFTFIG-1/Fig1-3-1.eps}$

Then angle $\theta$ is positive if you measure counter clockwise from the positive $x$

-axis.

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

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.

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

**Figure 1.13:** Trig function
$\includegraphics[width=3.5cm]{SOFTFIG-1/Fig1-3-2.eps}$

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 $\sin{\frac{\pi}{3}}$ .

**Figure 1.14:** Example1-5-1
$\includegraphics[width=3.5cm]{SOFTFIG-1/trigvalue_gr4.eps}$

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

**Figure 1.15:** Example1-5-2
$\includegraphics[width=3.5cm]{SOFTFIG-1/trigvalue_gr7.eps}$

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

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

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

**Figure 1.16:** Example1-5-3
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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

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

**Figure 1.17:** Example1-5-4
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Exercise 1..5 Find the value of the following.
1. $\displaystyle{\cos{\frac{5\pi}{4}}}$ 2. $\displaystyle{\sin(-\frac{5\pi}{6})}$ 3. $\displaystyle{\tan(-\frac{\pi}{3})}$

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

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

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

**Figure 1.18:** Exercise1-5-1
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**Figure 1.19:** Exercise1-5-2
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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

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

**Figure 1.20:** Exercise1-5-3
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Basic Trig Identities For all $\alpha,\beta$ ,
1. $\displaystyle{\cos^2{\alpha} + \sin^2{\alpha} = 1}$
2. $\displaystyle{\sin(\alpha \pm \beta) = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}}$
3. $\displaystyle{\cos(\alpha \pm \beta) = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}}$

**Figure 1.21:** addition-law
$\includegraphics[width=5cm]{SOFTFIG-1/addition-law.eps}$

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 $\cos(-\beta) = \cos{\beta}, \sin(-\beta) = -\sin{\beta}$ .

$\displaystyle \cos{\frac{5\pi}{4}}$	$\displaystyle =$	$\displaystyle \cos(\pi + \frac{\pi}{4}) = \cos \pi \cos\frac{\pi}{4} - \sin{\pi}\sin{\frac{\pi}{4}}$
	$\displaystyle =$	$\displaystyle (-1)\cdot \frac{\sqrt{2}}{2} - 0\cdot \frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}} \ensuremath{ \blacksquare}$

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

$\frac{5\pi}{6}$ $\frac{5\pi}{6} = \frac{\pi + 4\pi}{6} = \frac{\pi}{6} + \frac{2\pi}{3}$

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

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

$\displaystyle \sin(\frac{7\pi}{12})$	$\displaystyle =$	$\displaystyle \sin(\frac{\pi}{4}+\frac{\pi}{3}) = \sin{\frac{\pi}{4}}\cos{\frac{\pi}{3}} + \cos{\frac{\pi}{4}}\sin{\frac{\pi}{3}}$
	$\displaystyle =$	$\displaystyle \frac{\sqrt{2}}{2}\cdot \frac{1}{2} + \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2} + \sqrt{6}}{4} \ensuremath{ \blacksquare}$

$\frac{7\pi}{12}$ $\frac{7\pi}{12}$ is equivalent to $105^\circ$ and $105^\circ = 60^\circ + 45^\circ$

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

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

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

Half-angle formulas $\cos^{2}{\frac{x}{2}} = \frac{1 + \cos{x}}{2}$
$\sin^{2}{\frac{x}{2}} = \frac{1 - \cos{x}}{2}$