Prove that arcsin(1/2) + arcsin(1/3) = arcsin( (2√2+√3)/6 )

[hndalama]

Prove that arcsin(1/2) + arcsin(1/3) = arcsin( (2√2+√3)/6 )

I don't know where to start.

 

[stapel]

Prove that arcsin(1/2) + arcsin(1/3) = arcsin( (2√2+√3)/6 )

I don't know where to start.

Let arcsin(1/2) = x and arcsin(1/3) = y. Then, dealing with just the left-hand side and taking the sine of that side, we have sin(x + y). What identity can you apply to this? To what expression does "sin(x + y)" evaluate, after application of this identity? Where might this lead?

 

[hndalama]

Let arcsin(1/2) = x and arcsin(1/3) = y. Then, dealing with just the left-hand side and taking the sine of that side, we have sin(x + y). What identity can you apply to this? To what expression does "sin(x + y)" evaluate, after application of this identity? Where might this lead?

but isn't sin(x+y) ≠ arcsin(1/2) + arcsin (1/3)

and i think the identity you're referring to is sin(x + y) = sinxcosy + cosxsiny , but this looks like we have moved further from determining arcsin( (2√2+√3)/6 ). Where is this leading?

 

[Deleted member 4993]

but isn't sin(x+y) ≠ arcsin(1/2) + arcsin (1/3)

and i think the identity you're referring to is sin(x + y) = sinxcosy + cosxsiny , but this looks like we have moved further from determining arcsin( (2√2+√3)/6 ). Where is this leading?

Sheeesh.... are you using pencil and paper or just staring at the screen!!

Prove that arcsin(1/2) + arcsin(1/3) = arcsin( (2√2+√3)/6 )

I don't know where to start.

Let arcsin(1/2) = x and arcsin(1/3) = y. Then, dealing with just the left-hand side and taking the sine of that side, we have sin(x + y). What identity can you apply to this? To what expression does "sin(x + y)" evaluate, after application of this identity? Where might this lead? :wink:

sin^-1(1/2) = x → sin(x) = 1/2 & cos(x) = √3/2 and

sin^-1(1/3) = y → sin(y) = 1/3 & cos(y) = 2√2/3

sin[arcsin(1/2) + arcsin(1/3)] = sin(x + y) = ...... continue......

 

[stapel]

but isn't sin(x+y) ≠ arcsin(1/2) + arcsin (1/3)

I don't think anybody is saying that these are equal. Instead, I said to simplify by using the given substitution, and then to take the sine of the resulting expression.

and i think the identity you're referring to is sin(x + y) = sinxcosy + cosxsiny , but this looks like we have moved further from determining arcsin( (2√2+√3)/6 ). Where is this leading?

What did you get when you applied the identity? What did you get when you combined the two fractional expressions into one? What were your thoughts when you compared this with the right-hand side of what they gave you?

 

[hndalama]

sinxcosy + cosxsiny
1/2 * 2√2/3 +√3/2 * 1/3
(2√2 + √3)/6
arcsin (2√2 + √3)/6

Thanks guys