0.3: Trigonometry

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The right triangle

Trigonometric functions

$righttriangle2.jpg$

Usually in a triangle when there is no chance of ambiguity, the sides opposite to each vertex have length denoted by the corresponding small letter.

$righttriangle3.jpg$

In the diagram, the triangle $ABC$ has a right angle at $B.$

\[\sin (A)= \frac{\mbox {length of the opposite side}}{ \mbox{ length of the hypotenuse}}=\frac{a}{b}.\]
\[\cos (A)= \frac{\mbox {length of the adjacent side}}{ \mbox{ length of the hypotenuse}}=\frac{c}{b}.\]
\[\tan (A)= \frac{\mbox {length of the opposite side}}{ \mbox{ length of the length of the adjacent side}} =\frac{a}{c}= \frac{\sin (A)}{\cos (A)}.\]
The remanining functions can be considered to be reciprocal functions.
\[\csc (A)=\frac{1}{\sin (A)},\]
\[\sec (A)=\frac{1}{\cos (A)},\]
\[\cot (A)=\frac{1}{\tan (A)}.\]

Solving the general triangle

Rules

Consider the triangle $ABC$ in the diagram:

$triangle2.jpg$

1. The sine law:
\[\frac{a}{ \sin (A)}=\frac{b}{ \sin (B)}=\frac{c}{ \sin (C)}\]
2. The cosine law:
\[a^2=b^2+c^2-2bc \cos (A)\] or equivalently,
\[ \cos (A)= \frac{b^2+c^2-a^2}{2bc}\]
3. The area of the triangle $ABC$ is $\frac{1}{2} bc \sin (A).$
4. If the perimeter of the triangle $= 2s= a+b+c,$ then \[\sin (A)= \frac{2}{bc} \sqrt{s(s-a)(s-b)(s-c)}\]
and the area of the triangle is $\sqrt{s(s-a)(s-b)(s-c)}$.

Unit Circle

Definition

Unit circle is circle with center at the origin $(0,0)$ and radius $1$.
Unit circle has the equation $x^2+y^2=1.$

$circle1.jpg$

The radian measure of an angle is equal to the numerical value of the length of the arc of the unit circle from the point
$(1,0)$ to the point $(x,y).$

$circle2.jpg$

Thus, $\mbox{length of the arc } AP=\alpha \mbox{ radian units}.$
Hence we obtain the conversion from degrees to radians and vice-versa:\[\pi \,{\rm radians}=180^{\circ}\]

Definition of the trigonometric functions

There are two ways to define trigonometric functions:

Definition

Using triangles or circles.

1. The definition using triangles is too restrictive in that it treats only angles between $0$ degree and $90$ degrees.

2. The definition using circles, on the other hand, leads to a much more general definition, since the angle can now take on any real value whatsoever.

$circle2.jpg$

In the diagram above, $A(1,0)$ is the fixed point from which measurements begin. Angles and arc lengths are measured from $A$ in a counter-clockwise direction.

If $P(x,y)$ is any point on the circumference of the unit circle, then if the arc length of $AP= \alpha$ units, then $\angle AOP = \alpha$ radians, we define \[\sin (\alpha)=x \mbox{ and} \cos (\alpha)=y.\]

Further, we define \[\tan (\alpha)=\frac{\sin (\alpha)}{\cos (\alpha)} \]
\[\cot (\alpha)=\frac{\cos (\alpha)}{\sin (\alpha)}, \]
\[\sec (\alpha)=\frac{1}{\cos (\alpha)}, \]
\[\csc (\alpha)=\frac{1}{\sin (\alpha)}. \]

Some important relationships on unit circle

Rules

\[\left.\begin{array}{c}
\sin (\pi- \alpha)=\sin (\alpha).\\
\cos (\pi- \alpha)=-\cos (\alpha).\\
\tan (\pi- \alpha)=-\tan (\alpha).
\end{array}
\right\} \mbox{ Quadrant II}\]

\[\left.\begin{array}{c}
\sin (\pi+\alpha)=-\sin (\alpha).\\
\cos (\pi+ \alpha)=-\cos (\alpha).\\
\tan (\pi+ \alpha)=\tan (\alpha).
\end{array}
\right\} \mbox{ Quadrant III}\]

\[\left.\begin{array}{c}
\sin (2\pi- \alpha))=\sin (-\alpha)=-\sin (\alpha).\\
\cos (2\pi- \alpha)=\cos (-\alpha=\cos (\alpha).\\
\tan (2\pi- \alpha)=\tan (-\alpha)=-\tan (\alpha).
\end{array}
\right\} \mbox{ Quadrant IV}\]

\[\sin (\pi/2-\alpha)=\cos (\alpha).\]
\[\cos (\pi/2-\alpha)=\sin (\alpha).\]

Basic identities

\begin{eqnarray} \sin^2(A)+\cos^2(A)& =& 1.\\
1+\tan^2(A)&=&\sec^2(A).\\
1+\cot^2(A)&=&\csc^2(A).\\
\cos(A+B)&=&\cos(A)\cos(B)-\sin(A) \sin(B).\\
\cos(A-B)&=&\cos(A)\cos(B)+\sin(A) \sin(B).\\
\sin(A+B)&=&\sin(A)\cos(B)+\cos(A) \sin(B).\\
\sin(A-B)&=&\sin(A)\cos(B)-\cos(A) \sin(B).\\
\sin(2A)&=&2 \sin(A) \cos (A).\\
\cos(2A)&=& \cos^2(A)- \sin^2(A)\\
&=& 2 \cos^2(A)-1\\
&=& 1-2 \sin^2(A).\\
\tan(A+B)&= &\displaystyle\frac{\tan (A) + \tan (B)}{1-\tan (A) \tan (B)}.\\
\tan(A-B)&= &\displaystyle\frac{\tan (A) - \tan (B)}{1+\tan (A) \tan (B)}.\\
\tan(2A)&= &\displaystyle\frac{2\tan (A) }{1-\tan^2 (A) }.\\
\end{eqnarray}