# Thread: simplying trig expression

1. ## simplying trig expression

i have to simply the expression which is:

sin(tan-1(x))

now i'm not exactly where to start

what i've thought of so far is that since tan t = y/x which would be tan theta = sin/cos, tan-1 would be cos/sin but that is what cot theta is equal to.

i'm quite stumped as it is and don't exactly understand the problem.

thanks in advance

2. Originally Posted by deltemis
i have to simply the expression which is:

sin(tan-1(x))

now i'm not exactly where to start

what i've thought of so far is that since tan t = y/x which would be tan theta = sin/cos, tan-1 would be cos/sin but that is what cot theta is equal to.

i'm quite stumped as it is and don't exactly understand the problem.

thanks in advance
Let $\displaystyle \theta=\tan^{-1}x$.

Then if you construct a [right] triangle containing $\displaystyle \theta$, the side opposite of $\displaystyle \theta$ will have a length of x, the side adjacent to $\displaystyle \theta$ will have a length of 1, and the hypotenuse of the triangle will have a length of $\displaystyle \sqrt{x^2+1}$.

Therefore, $\displaystyle \sin\left(\tan^{-1}x\right)=\sin\theta=\frac{\text{opp}}{\text{hyp} }=\frac{x}{\sqrt{1+x^2}}$

Does this make sense?

3. thanks chris.

i keep forgetting the basics of trig and that is to work w/ triangles