Thread: Puzzling Inverse Trig Function solution

1. Puzzling Inverse Trig Function solution

Really hung up on this inverse trig function problem's solution. Actually I think my difficulty is in the algebra involved (not the actual trig), because; sin(arcsin(x)) = x, so really I'm wondering why (5/13) + pi/4 = 17sqrt2/26. Can anyone explain to me why that is in detail?

if you can't see the image, the problem is:

sin ( arcsin (5/13) + pi/4) = 17 sqrt 2 / 26

WHY?

2. Re: Puzzling Inverse Trig Function solution

let $\theta = \arcsin\left(\dfrac{5}{13}\right)$

$\sin\left(\theta + \dfrac{\pi}{4}\right) = \sin{\theta}\cos\left(\dfrac{\pi}{4}\right) + \cos{\theta}\sin\left(\dfrac{\pi}{4}\right)$

$\sin\left(\theta + \dfrac{\pi}{4}\right) = \dfrac{5}{13} \cdot \dfrac{\sqrt{2}}{2} + \dfrac{12}{13} \cdot \dfrac{\sqrt{2}}{2} = \dfrac{17\sqrt{2}}{26}$

of course, this is only true if $\theta$ is a quadrant I angle

3. Re: Puzzling Inverse Trig Function solution

Originally Posted by blakesgrave
Really hung up on this inverse trig function problem's solution. Actually I think my difficulty is in the algebra involved (not the actual trig), because; sin(arcsin(x)) = x, so really I'm wondering why (5/13) + pi/4 = 17sqrt2/26. Can anyone explain to me why that is in detail?
sin ( arcsin (5/13) + pi/4) = 17 sqrt 2 / 26
WHY?
Let $\theta=\arcsin\left(\frac{5}{13}\right)$ then

\begin{align*}\sin\left(\theta+\frac{\pi}{4}\right )&=\sin(\theta)\cos\left(\frac{\pi}{4}\right)+\sin \left(\frac{\pi}{4}\right)\cos(\theta)\\&=\sin( \theta)\left(\frac{\sqrt2}{2}\right)+\left(\frac{ \sqrt2}{2}\right)\cos(\theta)\\&=\left(\frac{12}{1 3}\right)\left(\frac{\sqrt2}{2}\right) +\left(\frac{5}{13}\right) \left(\frac{\sqrt2}{2} \right) \end{align*}

4. Re: Puzzling Inverse Trig Function solution

Ok, that all makes sense but now I'm wondering why

sin(θ + pi/4) = sinθcos(pi/4) + cosθ sin(pi/4)

where did cos come from?

5. Re: Puzzling Inverse Trig Function solution

Originally Posted by blakesgrave
Ok, that all makes sense but now I'm wondering why
sin(θ + pi/4) = sinθcos(pi/4) + cosθ sin(pi/4) where did cos come from?
$\sin(A\pm B)=\sin(A)\cos(B)\pm\sin(B)\cos(A)$ and $\cos(A\pm B)=\cos(A)\cos(B)\mp\sin(A)\sin(B)$