TRIGONOMETRIC RATIOS
Behaviour of trigonometric functions sin \(\theta\), cos\(\theta\) and tan\(\theta\) as changes from 00 to 900. i.e., \({0^0} \leqslant \theta \leqslant {90^0}\)
Case i: For
Let OX be the initial position of the rotating line. Take any point P on this line at a distance r units form the origin. Then the coordinates of the point Pare \((r,\theta)\) . From Fig note that OP makes an angle \(\theta=0^0\) with the X-axis. This means that x = r and y = 0. Then by definition
\(\cos ec\theta = \cos ec{0^0} = \frac{r}{y} = \frac{r}{0}\) (This ratio is not defined)
\(\sec \theta = \sec {0^0} = \frac{r}{x} = \frac{r}{r} = 1\)
\(\cot \theta = \cot {0^0} = \frac{x}{y} = \frac{r}{0}\) (This ratio is also not defined)
Case(ii) : For \(\theta = {30^0}or\frac{{{\mu ^c}}}{6}and\theta = {60^0}or\frac{{{\mu ^c}}}{3}\)
To obtain the values of trigonometrical ratios of \(\theta = {30^0}and{60^0},\) and 600 .consider the equvilateral triangle ABC of side 2a. Draw the perpendicular AD from the vertex A to BC. Then AD bisects side BC and the angle A.
\(\therefore BD = DC = a;\) \(\angle BAD = \angle CAD = {30^0}\)
We know that the height of an equilateral triangle with side 2a is equal to \(a\sqrt 3 \) . Then from the right angled \(\Delta ADB\) .
Case iii: For \(\theta = {45^0}\) or \(\frac{{{\pi ^C}}}{4}\)
To obtain the values of the six trigonometrical ratios corresponding to \(\angle B = {90^0}\) .Consider the \(\Delta ABC\) with AB = BC = a and \(\angle B=90^0\). Then it follows that \(\angle A = \angle C = {45^0}\).
From pythogorous theorem. \(A{B^2} + B{C^2} = A{C^2}\) and
\(AC = \sqrt {A{B^2} + B{C^2}} = a\sqrt 2 \)
Then \(\sin {45^0} = \frac{{AB}}{{AC}} = \frac{a}{{a\sqrt 2 }} = \frac{1}{{\sqrt 2 }};\)
\(\begin{gathered}
\cos {45^0} = \frac{{AB}}{{AC}} = \frac{a}{{a\sqrt 2 }} = \frac{1}{{\sqrt 2 }}; \hfill \\
\tan {45^0} = \frac{{BC}}{{AB}} = \frac{a}{a} = 1 \hfill \\
\cos ec{45^0} = \frac{{AC}}{{BC}} = \frac{{a\sqrt 2 }}{a} = \sqrt 2 ; \hfill \\
\sec {45^0} = \frac{{AC}}{{AB}} = \frac{{a\sqrt 2 }}{a} = \sqrt 2 ; \hfill \\
\cot {45^0} = \frac{{AB}}{{BC}} = \frac{a}{a} = 1 \hfill \\
\end{gathered} \)
Note : Observe here that instead of considering \(\angle A = \theta = {45^0},\) if we consider \(\angle C = \theta = {45^0}\) , we get same values for all the six ratios considered above.
Then \(\sin {90^0} = \frac{y}{r} = \frac{r}{r} = 1\) \(\cos {90^0} = \frac{x}{r} = \frac{0}{r} = 0;\)
\(\tan {90^0} = \frac{y}{x} = \frac{y}{0}\)(undefined)
\(\cos ec{90^0} = \frac{r}{y} = \frac{r}{r} = 1\) (undefined) \(\sec {90^0} = \frac{r}{x} = \frac{r}{0}\)
\(\cot {90^0} = \frac{x}{y} = \frac{0}{y} = 0\)