# How are trigonometric ratios function of interior angles in a right angled triangle?

How can one assume that the ratio altitude/hypotenuse is a function of angle. For a general right-angled triangle--->Let:

Hypotenuse$=c$

Altitude$=a$

Base$=b$

and angle opposite to altitude$=x$ .

Then by Pythagoras Theorem- $a^2+b^2=c^2 \qquad(1)$

and for a fixed value of angle $x$ ratio of sides is constant,therefore- $a/c=S$, $b/c=C$ or $a=cS$ , $b=cC$ putting these values in equation(1): $$(cS)^2+(cC)^2=c^2 \qquad S^2+C^2=1 \qquad(2)$$ Where $S$ and $C$ are some constants. My question is that how can I relate $S$ and $C$ with angle $x$.Do I simply write sin(x)=S,cos(x)=C,which according to me is just a declaration that S is a function of x.What is the real definition and how can it be derived?

• Usually, what you call perpendicular, is called altitude. Yes S and C are functions of x. That's the whole point. The only problem is that this only works when $0 \lt x^\circ \lt 90^\circ$. You need to do something else to define sine and cosine for any angle. – steven gregory Apr 10 '16 at 12:31
• Usually, what we call a number and use for counting could be replaced by sticks and bones..... – Ayush Dwivedi Apr 10 '16 at 14:04

## 1 Answer

By the definition of trigonometric functions you have: $$S=\frac{a}{c}=\frac{1}{\sin x} \qquad C=\frac{b}{c}=\frac{1}{\tan x}$$

so you have simply proved the identity:

$$\frac{1}{\sin^2 x}=\left(1+\frac{1}{\tan^2 x} \right)$$ or $$\mbox{cosec}^2 x=1+\mbox{cotan}^2 x$$