Limits of the Values of Trigonometrical Functions

The trigonometrical function explained before has some limiting values known as values of trigonometrical functions, which depend on the ratios’ value. Let’s consider a triangle OPM.

In the $ \displaystyle \Delta OPM$ it is obvious that none of the sides OM or PM can be equal to the hypotenuse OP i.e. both x and y are $ \displaystyle <r$.

In particular cases, when the radius vector OP, while moving in the clockwise direction tends to coincide with OX or while moving in the counter clockwise direction tends to coincide with OY, then x and y may be numerically equal to r but in no circumstances either x or y will be $ \displaystyle >r$. Hence x and y are always $ \displaystyle \le r$.

$ \displaystyle \begin{array}{l}\because r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\\Rightarrow \left| x \right|\le r\text{ and }\left| y \right|\le r\\\text{therefore,}\end{array}$

$\displaystyle \begin{array}{l}\text{(i) }\left| {\sin \theta } \right|=\left| {\dfrac{y}{r}} \right|=\dfrac{{\left| y \right|}}{{\left| r \right|}}=\dfrac{{\left| y \right|}}{r}\le 1\\\therefore -1\le \sin \theta \le 1\\\text{(ii) }\left| {\cos \theta } \right|=\left| {\dfrac{x}{r}} \right|=\dfrac{{\left| x \right|}}{{\left| r \right|}}=\dfrac{{\left| x \right|}}{r}\le 1\\\therefore -1\le \cos \theta \le 1\end{array}$

$ \displaystyle \begin{array}{l}\text{(iii) }\left| {\sec \theta } \right|=\left| {\dfrac{r}{x}} \right|=\dfrac{{\left| r \right|}}{{\left| x \right|}}=\dfrac{r}{{\left| x \right|}}\ge 1\\\therefore \sec \theta \le -1\text{ or }\sec \theta \ge 1\\\text{(iv) }\left| {\text{cosec}\theta } \right|=\left| {\dfrac{r}{y}} \right|=\dfrac{{\left| r \right|}}{{\left| y \right|}}=\dfrac{r}{{\left| y \right|}}\ge 1\\\therefore \text{cosec}\theta \le -1\text{ or }\left| {\text{sec}\theta } \right|\ge 1\end{array}$

$ \displaystyle {\text{(v) }\left| {\tan \theta } \right|=\left| {\dfrac{y}{x}} \right|=\dfrac{{\left| y \right|}}{{\left| x \right|}}}$ $ \displaystyle =$ any positive number

$ \displaystyle {\therefore -\infty \le \tan \theta \le \infty }$

$ \displaystyle {\text{(vi) }\left| {\cot \theta } \right|=\left| {\dfrac{x}{y}} \right|=\dfrac{{\left| x \right|}}{{\left| y \right|}}}$ $ \displaystyle =$ any positive number

$ \displaystyle {\therefore -\infty \le \tan \theta \le \infty }$

Therefore, the values of $ \displaystyle {\sin \theta }$ and $ \displaystyle {\cos \theta }$ will always lie between -1 and +1; $ \displaystyle -1\le \sin \theta \le 1$ and $ \displaystyle -1\le \cos \theta \le 1$.

In other words, $\displaystyle \left| {\sin \theta } \right|\le 1$ and $ \displaystyle \left| {\cos \theta } \right|\le 1$.

Since $ \displaystyle \text{cosec}\theta $ and $ \displaystyle \sec \theta $, respectively are the reciprocals of $\displaystyle \left| {\sin \theta } \right|\le 1$ and $ \displaystyle \left| {\cos \theta } \right|\le 1$, therefore their values (numerically) are always greater than or equal to 1. That means $ \displaystyle \text{cosec}\theta \text{, }\sec \theta \ge 1$ as also $ \displaystyle \le -1$ or $ \displaystyle \left| {\text{cosec}\theta } \right|\ge 1$ and $ \displaystyle \left| {\text{sec}\theta } \right|\ge 1$. But $ \displaystyle \tan \theta $ and $ \displaystyle {\cot \theta }$ may have any values; positive or negative.

We may verify the truth by looking at the trigonometrical ratios of some important angles (viz; $ \displaystyle {{0}^{\circ }},{{60}^{\circ }},{{90}^{\circ }},{{120}^{\circ }}\text{ etc}$)

Relation between the trigonometrical ratios of angle (Basic Formulae)

Let O be the vertex of an angle A, OX its initial ray and OP its terminal ray. Let P (x, y) be a point on the terminal ray.

Let PL be the perpendicular to OX’. Let OP = r then

Now,

$ \displaystyle \text{(i) }\sin A=\dfrac{y}{r}$

$ \displaystyle {\text{(ii) }\cos A=\dfrac{x}{r}}$

$ \displaystyle {\text{(iii) }\tan A=\dfrac{y}{x}\text{, }\cot A=\dfrac{x}{y}\text{,}}$ $ \displaystyle {\sec A=\dfrac{r}{x}\text{, cosec }A=\dfrac{r}{y}}$

$ \displaystyle {\text{1}\text{. }{{{\sin }}^{2}}A+{{{\cos }}^{2}}A=1}$

$ \displaystyle {\text{Proof: }{{{\sin }}^{2}}A+{{{\cos }}^{2}}A={{{\left( {\dfrac{y}{r}} \right)}}^{2}}+{{{\left( {\dfrac{x}{r}} \right)}}^{2}}}$ $ \displaystyle {=\dfrac{{{{y}^{2}}+{{x}^{2}}}}{{{{r}^{2}}}}=\dfrac{{{{r}^{2}}}}{{{{r}^{2}}}}=1}$

$ \displaystyle {\text{2}\text{. }{{{\sec }}^{2}}A-{{{\tan }}^{2}}A=1}$

$ \displaystyle {\text{Proof: }{{{\sec }}^{2}}A-{{{\tan }}^{2}}A={{{\left( {\dfrac{r}{x}} \right)}}^{2}}-{{{\left( {\dfrac{y}{x}} \right)}}^{2}}}$ $ \displaystyle {=\dfrac{{{{r}^{2}}-{{y}^{2}}}}{{{{x}^{2}}}}=\dfrac{{{{x}^{2}}}}{{{{x}^{2}}}}=1}$

$ \displaystyle {\text{3}\text{. cose}{{\text{c}}^{2}}A-{{{\cot }}^{2}}A=1}$

$ \displaystyle {\text{Proof: cose}{{\text{c}}^{2}}A-{{{\cot }}^{2}}A={{{\left( {\dfrac{r}{y}} \right)}}^{2}}-{{{\left( {\dfrac{x}{y}} \right)}}^{2}}}$ $ \displaystyle {=\dfrac{{{{r}^{2}}-{{x}^{2}}}}{{{{y}^{2}}}}=\dfrac{{{{y}^{2}}}}{{{{y}^{2}}}}=1}$

$ \displaystyle {\text{4}\text{. }\sin A=\dfrac{1}{{\text{cosec }A}},\text{ }\cos A=\dfrac{1}{{\sec A}}}$ and $ \displaystyle {\tan A=\dfrac{1}{{\cot A}}}$

Important Results

$ \displaystyle {\text{1}\text{. }{{{\sin }}^{2}}A+{{{\cos }}^{2}}A=1}$

$ \displaystyle {\text{2}\text{. }\sin A=\pm \sqrt{{1-{{{\cos }}^{2}}A}};}$ $ \displaystyle {\cos A=\pm \sqrt{{1-{{{\sin }}^{2}}A}}}$

positive or negative sign before the root depends on the quadrant in which A lies

$ \displaystyle {\text{3}\text{. }\tan A=\pm \sqrt{{{{{\sec }}^{2}}A-1}};}$ $ \displaystyle {\sec A=\pm \sqrt{{1+{{{\tan }}^{2}}A}}}$

$ \displaystyle {\text{4}\text{. }\cot A=\pm \sqrt{{\text{cose}{{\text{c}}^{2}}A-1}};}$ $ \displaystyle {\text{cosec }A=\pm \sqrt{{1+{{{\cot }}^{2}}A}}}$

$ \displaystyle {\text{5}\text{. }\sin A\times \text{cosec }A=1;}$ $ \displaystyle {\cos A\times \sec A=1;}$ $ \displaystyle {\tan A\times \cot A=1}$

Example 1:

Prove that $ \displaystyle \sqrt{{\dfrac{{1+\sin \theta }}{{1-\sin \theta }}}}=\sec \theta +\tan \theta $

Solution:

$ \displaystyle \begin{array}{l}\text{L}\text{.H}\text{.S}\text{. }=\sqrt{{\dfrac{{1+\sin \theta }}{{1-\sin \theta }}}}\\=\sqrt{{\dfrac{{1+\sin \theta }}{{1-\sin \theta }}\times \dfrac{{1+\sin \theta }}{{1+\sin \theta }}}}\\=\sqrt{{\dfrac{{{{{\left( {1+\sin \theta } \right)}}^{2}}}}{{1-{{{\sin }}^{2}}\theta }}}}\\=\dfrac{{1+\sin \theta }}{{\cos \theta }}\\=\dfrac{1}{{\cos \theta }}+\dfrac{{\sin \theta }}{{\cos \theta }}\\=\sec \theta +\tan \theta \end{array}$

Example 2:

Given $ \displaystyle 5\cos \theta +3=0$ and $ \displaystyle \theta $ is in the second quadrant; find the other trigonometrical ratios.

Solution:

From the given equation, we have $ \displaystyle \cos \theta =-\dfrac{3}{5}$. Also $ \displaystyle \theta $ is in the second quadrant. Therefore, $ \displaystyle \sin \theta $ will be positive and $ \displaystyle \tan \theta $ will be negative.

Now $ \displaystyle \begin{array}{l}{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta =1-\dfrac{9}{{25}}=\dfrac{{16}}{{25}}\\\therefore \sin \theta =\dfrac{4}{5}\\\text{Hence, }\tan \theta =\dfrac{4}{3}\text{ etc}\text{.}\end{array}$