10.1 The trig functions and 2-d problems  (Page 2/2)

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 ${0}^{\circ }$ ${30}^{\circ }$ ${45}^{\circ }$ ${60}^{\circ }$ ${90}^{\circ }$ ${180}^{\circ }$ $cos\theta$ 1 $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{2}$ 0 $-1$ $sin\theta$ 0 $\frac{1}{2}$ $\frac{1}{\sqrt{2}}$ $\frac{\sqrt{3}}{2}$ 1 0 $tan\theta$ 0 $\frac{1}{\sqrt{3}}$ 1 $\sqrt{3}$ $-$ 0

These values are useful when asked to solve a problem involving trig functions without using a calculator.

Each of the trigonometric functions has a reciprocal that has a special name. The three reciprocals are cosecant (or cosec), secant (or sec) and cotangent (or cot). These reciprocals are given below:

$\begin{array}{ccc}cosec\theta & =& \frac{1}{sin\theta }\\ sec\theta & =& \frac{1}{cos\theta }\\ cot\theta & =& \frac{1}{tan\theta }\end{array}$

We can also define these reciprocals for any right angled triangle:

$\begin{array}{ccc}\hfill cosec\theta & =& \frac{\mathrm{hypotenuse}}{\mathrm{opposite}}\hfill \\ \hfill sec\theta & =& \frac{\mathrm{hypotenuse}}{\mathrm{adjacent}}\hfill \\ \hfill cot\theta & =& \frac{\mathrm{adjacent}}{\mathrm{opposite}}\hfill \end{array}$

Find the length of x in the following triangle.

1. In this case you have an angle ( ${50}^{\circ }$ ), the opposite side and the hypotenuse.

So you should use $sin$

$sin{50}^{\circ }=\frac{x}{100}$
2. $⇒x=100×sin{50}^{\circ }$
3. Use the sin button on your calculator

$⇒x=76.6\mathrm{m}$

Find the value of $\theta$ in the following triangle.

1. In this case you have the opposite side and the hypotenuse to the angle $\theta$ .

So you should use $tan$

$tan\theta =\frac{50}{100}$
2. $tan\theta =0.5$
3. Since you are finding the angle ,

use ${tan}^{-1}$ on your calculator

Don't forget to set your calculator to `deg' mode!

$\theta =26.{6}^{\circ }$

In the previous example we used ${\mathrm{tan}}^{-1}$ . This is simply the inverse of the tan function. Sin and cos also have inverses. All this means is that we want to find the angle that makes the expression true and so we must move the tan (or sin or cos) to the other side of the equals sign and leave the angle where it is. Sometimes the reciprocal trigonometric functions are also referred to as the 'inverse trigonometric functions'. You should note, however that ${\mathrm{tan}}^{-1}$ and $\mathrm{cot}$ are definitely NOT the same thing.

The following videos provide a summary of what you have learnt so far.

Finding lengths

Find the length of the sides marked with letters. Give answers correct to 2 decimal places.

Two-dimensional problems

We can use the trig functions to solve problems in two dimensions that involve right angled triangles. For example if you are given a quadrilateral and asked to find the one of the angles, you can construct a right angled triangle and use the trig functions to solve for the angle. This will become clearer after working through the following example.

Let ABCD be a trapezium with $\mathrm{AB}=4\phantom{\rule{1pt}{0ex}}\mathrm{cm}$ , $\mathrm{CD}=6\phantom{\rule{1pt}{0ex}}\mathrm{cm}$ , $\mathrm{BC}=5\phantom{\rule{1pt}{0ex}}\mathrm{cm}$ and $\mathrm{AD}=5\phantom{\rule{1pt}{0ex}}\mathrm{cm}$ . Point E on diagonal AC divides the diagonal such that $\mathrm{AE}=3\phantom{\rule{1pt}{0ex}}\mathrm{cm}$ . Find $A\stackrel{^}{B}C$ .

1. We draw a diagram and construct right angled triangles to help us visualize the problem.
2. We will use triangle ABE and triangle BEC to get the two angles, and then we will add these two angles together to find the angle we want.
3. We use sin for both triangles since we have the hypotenuse and the opposite side.
4. In triangle ABE we find:
$\begin{array}{ccc}\hfill sin\left(A\stackrel{^}{B}E\right)& =& \frac{\mathrm{opp}}{\mathrm{hyp}}\hfill \\ \hfill sin\left(A\stackrel{^}{B}E\right)& =& \frac{3}{4}\hfill \\ A\stackrel{^}{B}E& =& {sin}^{-1}\left(\frac{3}{4}\right)\hfill \\ A\stackrel{^}{B}E& =& {48,59}^{\circ }\hfill \end{array}$
We use the theorem of Pythagoras to find $\mathrm{EC}=4,4\phantom{\rule{1pt}{0ex}}\mathrm{cm}$ . In triangle BEC we find:
$\begin{array}{ccc}\hfill sin\left(C\stackrel{^}{B}E\right)& =& \frac{\mathrm{opp}}{\mathrm{hyp}}\hfill \\ \hfill sin\left(C\stackrel{^}{B}E\right)& =& \frac{4,4}{5}\hfill \\ A\stackrel{^}{B}E& =& {sin}^{-1}\left(\frac{4,4}{5}\right)\hfill \\ C\stackrel{^}{B}E& =& {61,64}^{\circ }\hfill \end{array}$
5. We add the two angles together to get: ${48,59}^{\circ }+{61,64}^{\circ }={110,23}^{\circ }$

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