Question

Please explain the meaning of sin^-1, cos^-1, tan^-1 in terms of sin, cos,tan respectively

Answer 1

Let us take a right angle triangle ABC, with right angle at B.

Sine of A is defined as BC / AC  = opposite side to angle A / hypotenuse

$Sin A = \frac{BC}{AC}$       example:  Sin 30 deg  = 1/2

We want to know the value of angle A, but we may not directly know it. Then we can calculate its sine value from other quantities in some way, and then find its value as:

$sin^{-1} 1/2 = 30\ deg$

Thus  we define $sin^{-1} BC/AC$ as equal to angle A.

Same way, we know $Cosine\ C = \frac{BC}{AC}$
so we define $Cos^{-1} \frac{BC}{AC} = angle C$

Similarly we know the  tangent $tan C = \frac{AB}{BC}$
We want to find the angle C, when we are given AB and BC. So we define a function
$tan^{-1} \frac{AB}{BC}\ \ as\ angle\ C$


So sine cosine, tangent of angles give ratio of sides in a right angle triangle. The inverse functions give the values of angles from ratio of sides.