Inverse Trig Functions Composite functions of Csc, Sec, And Cot

Question

Ok guys.. I'm trying to get prepared for my test tomorrow and I did numerous exercises. But I stumbled upon one of the "types" of exercises. Which is a composite function in $\csc$ and $\sec$. For example, it would give you this:

1. $\csc\left[\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)\right]$
2. $\sec\left[\sin^{-1}\left(-\dfrac{2}{5}\right)\right]$

So those types of questions I need to understand more on. Help would be very much appreciated!(Try to relate to Unit Circle Please..)

Answer 1

Firstly note that $\cos^{-1}\frac{\sqrt{3}}{2}$ must be in the first quadrant so any other trigonometric calculations must be positive.

You then have two possible approaches:

Option 1

Let $x=\cos^{-1}\frac{\sqrt{3}}{2}$ so you have that $\cos x=\frac{\sqrt{3}}{2}$. Draw the triangle:

Calculate the length of the other side (using Pythagoras), it is of length 1.

Then calculate $\csc x=\frac{1}{\sin x}=\frac{1}{\frac{1}{2}}=2$

Option 2

Note that we are trying to evaluate $\csc x=\frac{1}{\sin x}$

Let $x=\cos^{-1}\frac{\sqrt{3}}{2}$ so you have that $\cos x=\frac{\sqrt{3}}{2}$.

Use $\sin^2 x + \cos^2 x = 1$ to get: $\sin x = \sqrt{1-\cos^2 x}=\sqrt{1-\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{1-\frac34}=\sqrt{\frac14}=\frac12$

Evalute $\csc x=\frac{1}{\sin x}=\frac{1}{\frac12}=2$

Comparison of Options

Both use the same underlying calculation. The first is easier numerically/algebraically but requires you to draw a diagram and remember all your ratios correctly. The second requires stronger algebraic skills and remembering that $\sin^2 x+\cos^2 x=1$ as well as which reciprocal functions are which.

Answer 2

edited to make more clear

If you can do one of these problems you can do any of them, for example let us do your second problem. The key here is that the inverse trig on the inside gives you an ANGLE. If that angle happens to be in the first quadrant, (which it will be in all of these types of problems) then we can easily compute the value using right triangles.

set $\theta = \sin^{-1}{2/5}$

Draw a right triangle with the angle $\theta$ in it, then by the definition of the angle $\theta$ we have $\sin(\theta) = \frac{2}{5}$. Now the definition of $sin$ in a right triangle is opposite side over adjacent side. So set the side opposite $\theta$ to 2 and the hypotenuse to 5. Use the Pythagorean theorem to find the 3rd side. Now the secant of $\theta$ is by definition the ratio of the hypotenuse over adjacent