Trig Crazy

**redpanda11** · October 2nd 2007, 06:23 PM

Evaluate:
a) tan(cos^-1(1/3)

b) Solve the pair of equations cosx=-1/2 and tanx=root3 for 0<-x<-2pi

**Soroban** · October 2nd 2007, 07:11 PM

Hello, redpanda11!

a) Evaluate: . $\tan\left[\cos^{-1}\!\left(\frac{1}{3}\right)\right]$

Recall that an inverse trig function is an angle.

Let: $\theta = \cos^{-1}\!\left(\frac{1}{3}\right)\quad\Rightarrow\quad\ cos\theta \:=\:\frac{1}{3} \:=\:\frac{adj}{hyp}$

$\theta$ is in a right triangle with: $adj = 1,\:hyp = 3$
Using Pythagorus, we find that: . $opp \:= \:\sqrt{8} \:=\:2\sqrt{2}$
Then: . $\tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{2\sqrt{2}}{1} \:=\:2\sqrt{2}\quad\Rightarrow\quad \theta \:=\:\tan^{-1}\left(2\sqrt{2}\right)<br />$

Therefore, we have: . $\tan\left[\tan^{-1}\!\left(2\sqrt{2}\right)\right] \;=\;\boxed{2\sqrt{2}}$

b) Solve the pair of equations: $\cos x = \text{-}\frac{1}{2}$ and $\tan x = \sqrt{3}$ . for $0 \leq x \leq 2\pi$

Cosine is negative in Quadrants 2 and 3.
Tangent is positive in Quadrants 1 and 3.
. . Hence, $x$ is in Quadrant 3.

Multiply the two equations: . $(\cos x)(\tan x) \;=\;\left(-\frac{1}{2}\right)\left(\sqrt{3}\right)$

Then we have: . $\cos x\cdot\frac{\sin x}{\cos x} \;=\;-\frac{\sqrt{3}}{2}\quad\Rightarrow\quad\sin x \:=\:-\frac{\sqrt{3}}{2}$

Therefore: . $\boxed{x \;=\;\frac{4\pi}{3}}$

**DivideBy0** · October 2nd 2007, 10:23 PM

Ooh multiplying equations... never thought of it, but it does make sense

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