sin[(inverse cos)(1/2)+(inverse sin)(3/5)]

Question

anonymous · Answer

$$\sin (\cos^{-1} (1/2)+\sin^{-1} (3/5))$$

loser66 · Answer

sin(a+b)=?

anonymous · Answer

sin(a+b)= sinAcosB+cosAsinB
but i'm not sure how to use that to answer this

loser66 · Answer

ok, take step, just do as usual. A = cos^-1(1/2) and B =sin^-1 (3/5) , expand it, let see what we have

loser66 · Answer

$$sin(cos^{-1}(\frac{1}{2}))*cos(sin^{-1}(\frac{3}{5}))+cos(cos^{-1}(\frac{1}{2})*sin(sin^{-1}(\frac{3}{5}))$$

loser66 · Answer

got this part?

anonymous · Answer

yup, i'm with you so far!

loser66 · Answer

$cos^{-1}(\frac{1}{2})=\frac{\pi}{3}$, right? so the first term is sin($\frac{\pi}{3})=\frac{\sqrt{3}}{2}$got it?

anonymous · Answer

yeah! for the cos, it's on the unit circle, and for the sin, you make a triangle

loser66 · Answer

I don't know what you mean. hehehe, but it doesn't matter, as long as we meet each other, we aaaare fiiine, right?

anonymous · Answer

yup! thank you! i think i understand now :)

loser66 · Answer

ok, can you take it over by yourself? don't tell me a big NNNNOOO, I am dummy now, cannot work anymore, hehehe

anonymous · Answer

yeah! i think i got it! thanks for your help! :)

loser66 · Answer

hehehe.... glad to hear