TRIGONOMETRIC RATIOS

TRIGONOMETRIC RATIOS
Behaviour of trigonometric functions sin\(\theta\), cos\(\theta\) and tan\(\theta\) as \(\theta\) changes from 0⁰ to 90⁰. i.e., \({0^0} \leqslant \theta \leqslant {90^0}\)
Case i: For \(\theta=90^0\)
    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 \(\theta = {30^0}or\frac{{{\pi ^c}}}{6}and\theta = {60^0}or\frac{{{\pi ^c}}}{3}\)
    To obtain the values of trigonometrical ratios of  \(\theta=0^0\)  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 \(\Delta ADB\).

\( \begin{gathered} \sin {30^0} = \frac{{opp.side}}{{Hyp}} = \frac{{BD}}{{AB}} = \frac{a}{{2a}} = \frac{1}{2} \hfill \\ \cos {30^0} = \frac{{adj.side}}{{Hyp}} = \frac{{DA}}{{AB}} = \frac{{a\sqrt 3 }}{{2a}} = \frac{{\sqrt 3 }}{2} \hfill \\ \tan {30^0} = \frac{{opp.side}}{{adj.side}} = \frac{{BD}}{{AD}} = \frac{a}{{a\sqrt 3 }} = \frac{2}{{\sqrt 3 }} \hfill \\ \cos ec{30^0} = \frac{{AB}}{{BD}} = 2; \hfill \\ \sec {30^0} = \frac{{AB}}{{AD}} = \frac{2}{{\sqrt 3 }}and,\cot {30^0} = \frac{{AD}}{{BD}} = \sqrt 3 \hfill \\ \end{gathered} \)
\( \begin{gathered} \sin {60^0} = \frac{{AD}}{{AB}} = \frac{{a\sqrt 3 }}{{2a}} = \frac{{\sqrt 3 }}{2}; \hfill \\ \cos {60^0} = \frac{{BD}}{{AB}} = \frac{a}{{2a}} = \frac{1}{2}; \hfill \\ \tan {60^0} = \frac{{AD}}{{BD}} = \frac{{a\sqrt 3 }}{a} = \sqrt 3 ; \hfill \\ \cos ec{60^0} = \frac{{AB}}{{AD}} = \frac{2}{{\sqrt 3 }} \hfill \\ \sec {60^0} = \frac{{AB}}{{BD}} = 2; \hfill \\ \cot {60^0} = \frac{{BD}}{{AD}} = \frac{1}{{\sqrt 3 }} \hfill \\ \end{gathered} \)
Case iii: For \(\theta=45^0\) or  \(\frac{{{\pi ^C}}}{4}\)  

  To obtain the values of the six trigonometrical ratios corresponding to \(\theta=45^0\) .Consider the 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)
\(\cos ec{90^0} = \frac{r}{y} = \frac{r}{r} = 1\)        \(\sec {90^0} = \frac{r}{x} = \frac{r}{0}\) (undefined)
\(\cot {90^0} = \frac{x}{y} = \frac{0}{y} = 0\)
Some standard angles : 

\(f(-\theta)\) for all the values of ‘\(\theta\)’: 
  \(\begin{gathered} \sin \left( { - \theta } \right) = - \sin \theta ; \hfill \\ \cos ec\left( { - \theta } \right) = - \cos ec\theta \hfill \\ \cos \left( { - \theta } \right) = \cos \theta \hfill \\ \end{gathered} \)      
   \(\begin{gathered} \sec \left( { - \theta } \right) = \sec \theta \hfill \\ \tan \left( { - \theta } \right) = - \tan \theta ; \hfill \\ \cot \left( { - \theta } \right) = - \cot \theta \hfill \\ \end{gathered} \)     
sign will change for sin, cosec, tan and cot values. There is  no change in cos and sec values.