Combo Trigonometry

Thanks Mel. That test should be much easier now. All l need to do is see that the inverse trig is the ANGLE of the outer trig function. A reference triangle should be a good way to use that angle.

Hi HSC it is good to see you again:)

l would like to know how combo trig works when you use it with inverse trig. Conceptually that is.

i.e. cos(arctan(x)) or sin(arccos(x))

I will look at     cos(arctan(x))

First you must realize that   atan(x) is an angle.  

So to make it easier to think straight I will call it  \(\theta. \)   

Also there are restrictions on theta.    -pi/2 < theta <pi/2

so theta is in the 1st or the forth quadrant.

You are finding cos of theta and cos is positive in the first and the 4th quads so the answer will be positive.

If you were finding something else like sin(theta) then the answer could be plus or minus.

I would do it the same way but then I'd stick a  plus/minus in front of the final answer.

Anyway.. Now I will just look at the 1st quad outcomes.

\(tan\theta=x=\frac{x}{1}=\frac{opp}{adj}\)

I'd draw a right angle triangle displaying this info and find the hypotenuse using pythagoras's theorum.

Now you can see from the pic that

\(cos\theta=\frac{1}{\sqrt{1+x^2}}\\ so\\ cos[arctan(x)]=\frac{1}{\sqrt{1+x^2}}\\\)

Now you should be able to give the second one a go by yourself :)