inverse trig expressions: simplify sin[2sin^-1(1/2)] , etc

[youknowjojo]

can some one help me find the exact value of some of these expressions step by step:

sin[2sin^-1(1/2)]

cot^2[1/2tan^-1(4/3)]

cot[sec^-1(5/3)+pi/6]

P.S. i didnt know how to enter a pi symbol so i just spelled out "pi".

 

[soroban]

Re: Trigonometry help

Hello, youknowjojo!

First of all, inverse trig functions have an infinite number of values.
. . I will use the least positive value.

I remind myself that an inverse trig expression is an angle.
Then I baby-step through the reasoning . . .

\(\displaystyle \sin\left[2\cdot\sin^{-1}\left(\frac{1}{2}\right)\right]\)

Let \(\displaystyle \theta \,=\,\sin^{-1}\left(\frac{1}{2}\right)\;\;\Rightarrow\;\;\sin\theta\,=\,\frac{1}{2}\;\;\Rightarrow\;\;\theta\,=\,\frac{\pi}{6}\)

The problem becomes: \(\displaystyle \:\sin\left[2\cdot\frac{\pi}{6}\right] \:=\:\sin\left[\frac{\pi}{3}\right] \:=\:\frac{\sqrt{3}}{2}\)

\(\displaystyle \cot^2\left[\frac{1}{2}\cdot\tan^{-1}\left(\frac{4}{3}\right)\right]\)

This one is quite unpleasant.
I'll come back to it later . . . or let someone else explain it.

\(\displaystyle \cot\left[\sec^{-1}\left(\frac{5}{3}\right)\,+\,\frac{\pi}{6}\right]\)

This one is also unpleasant, but I'll muddle through it . . .

Let \(\displaystyle \theta\,=\,\sec^{-1}\left(\frac{5}{3}\right)\)
Then we have: \(\displaystyle \:\cot\left(\theta\,+\,\frac{\pi}{6}\right) \;=\;\L\frac{1}{\tan\left(\theta\,+\,\frac{\pi}{6}\right)} \;=\;\frac{1\,-\,\tan(\frac{\pi}{6})\cdot\tan\theta}{\tan\theta\,+\,\tan(\frac{\pi}{6})}\;\) [1]

Here's the baby-talk . . .
We have: \(\displaystyle \:\theta\,=\,\sec^{-1}\left(\frac{5}{3}\right)\;\;\Rightarrow\;\; \sec\theta\,=\,\frac{5}{3}\)
Angle \(\displaystyle \theta\) is in a right triangle with: \(\displaystyle \,hyp\,=\,5,\:adj\,=\,3\)
Using Pythagorus, we find that: \(\displaystyle \,opp\,=\,4\)
. . Hence: \(\displaystyle \fbox{\tan\theta\,=\,\frac{4}{3}}\)
We also know that: \(\displaystyle \fbox{\tan(\frac{\pi}{6}) \,=\,\frac{\sqrt{3}}{3}}\)

Substitute these into [1]: \(\displaystyle \L\:\frac{1\,-\,\left(\frac{\sqrt{3}}{3}\right)\left(\frac{4}{3}\right)}{\frac{4}{3}\,+\,\frac{\sqrt{3}}{3}}\) . . . now I'll simplify (ack!)


Multiply top and bottom by \(\displaystyle 9:\;\L\frac{9\,-\,4\sqrt{3}}{3(4\,+\,\sqrt{3})}\)

Rationalize: \(\displaystyle \L\:\frac{9\,-\,4\sqrt{3}}{3(4\,+\,\sqrt{3})}\,\cdot\,\frac{4\,-\,\sqrt{3}}{4\,-\,\sqrt{3}}\;=\;\frac{36\,-\,9\sqrt{3}\,-\,16\sqrt{3}\,+\,12}{3(16\,-\,4\sqrt{3}\,+\,4\sqrt{3}\,-\,3)}\)

Answer: \(\displaystyle \L\:\frac{48\,-\,25\sqrt{3}}{39}\)

. . . I need a nap . . .