sin(2arcSine(4/3)+arcCos(1/3))=?

Question

anonymous · Answer

$$\sin (2 Sin ^{-1} \frac{ 4 }{ 5 }+Cos ^{-1}\frac{ 1 }{ 5 })$$

anonymous · Answer

It's actually 4/5 and 1/5 not 3. and the question makes a distinction between sin(lower case) and Sin(upper case)

anonymous · Answer

Plug in sin^-1(4/5 and cos^-1(1/5) in to your calculator. Then multiply what you get for sin^-1 by 2 because there is a 2 in front of it and add it with whatever you get for cos^-1. Then just take the sine of this value in the brackets.

@arkgolucky

anonymous · Answer

I'm not supposed to use a calculator @genius12

anonymous · Answer

Well the you can use the Sum/Difference identity for sine.$$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$$

What is 'a' and what is 'b' in this case?  @ArkGoLucky

anonymous · Answer

is the answer $$\frac{ 48\sqrt{6} }{ 625 }$$

anonymous · Answer

@genius12

anonymous · Answer

I haven't done it lol. I'll do it.

dumbcow · Answer

to check your answer http://www.wolframalpha.com/input/?i=sin%282arcsin%284%2F5%29%2Barccos%281%2F5%29%29 no you are not correct

anonymous · Answer

And yes you aren't correct.

anonymous · Answer

so how do you do it

anonymous · Answer

You use a number of trig identities of sum/difference and products of cosines/sines. Which you can find online and I don't have the time atm to post them. @ArkGoLucky