# Trigonometry Problems [involves inverse functions]

• Mar 20th 2011, 09:56 AM
FallenStar117
Trigonometry Problems [involves inverse functions]
So I am having a bit of trouble studying for my trig test coming up.... (Headbang)
for the section on inverse trig functions I am a bit confused with some questions

ex: sin[29arctanx)+arcos(2x)]

ex: cos[arctan3 - arcsin(-0.5)]

I understand the basics such as remembering the domains of the different functions and what not... but these questions seem to stump me.

Can you maybe solve one but with a thorough description.... =) (Clapping)(Clapping)(Clapping)(Bow)(Bow)(Bow)
• Mar 20th 2011, 10:40 AM
skeeter
I assume the first expression has a typo, and should be

$\sin[2\arctan{x} + \arccos(2x)]
$

for the second expression ...

$\cos[\arctan(3) - \arcsin(-0.5)]$

let $\theta = \arctan(3)$ , and you should already know that $\arcsin(-0.5) = -\dfrac{\pi}{6}$

using the sum identity for cosine ...

$\cos\left(\theta + \dfrac{\pi}{6}\right) = \cos{\theta} \cdot \cos\left(\dfrac{\pi}{6}\right) - \sin{\theta} \cdot \sin\left(\dfrac{\pi}{6}\right)$

since $\tan{\theta} = 3$ , then $\cos{\theta} = \dfrac{1}{\sqrt{10}}$ and $\sin{\theta} = \dfrac{3}{\sqrt{10}}$

sub these values into the sum expression above ...

$\cos\left(\theta + \dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{10}} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{3}{\sqrt{10}} \cdot \dfrac{1}{2} = \dfrac{\sqrt{3} - 3}{2\sqrt{10}}$
• Mar 20th 2011, 06:38 PM
FallenStar117
So how is it you found the value of arctan(3)?
• Mar 20th 2011, 06:55 PM
skeeter
Quote:

Originally Posted by FallenStar117
So how is it you found the value of arctan(3)?

I didn't ... I said let $\theta = \arctan(3)$ , which also says $\tan{\theta} = 3$ ... I used this fact to calculate $\sin{\theta}$ and $\cos{\theta}$ which was necessary evaluate the original expression.
• Mar 21st 2011, 09:28 AM
FallenStar117
So how would I start solving the first question?
Is the idea the same as the second question...