# sin[arcsin(5/13) + arccos(3/5)] = ?I found it as 77/65, but my answer says 63/65. What did I do wrong?

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Use the identity: sin(A+B)=sinAcosB + cosAsinB

sinA=5/13 cosB=3/5

cosA=12/13 sinB=4/5

then:(5/13)(3/5) + (12/13)(4/5) = (15/65) + (48/65) = (63/65)

First of all, let's establish some steps:

sin x= f(x) and we know that f(x) is a value. For example,

sin (pi/6) = 1/2

Now, in order to find the inverse function, we'll have:

x=arcsin f(x). But, from the start, we know that x is an angle, so arcsin f(x) is an angle, too.

sin (pi/6) = 1/2, so (pi/6)=arcsin (1/2)

Let's analyze sin x= f(x) again and to put instead of x, the inverse function, namely, arcsin f(x).

sin [arcsin f(x)]=f(x)

Let's focus on our expression now!

Based on what I've said before, arcsin(5/13) + arccos(3/5) is nothing else but a sum between 2 angles, which are arcsin(5/13) and arccos(3/5).

We know the formula for adding 2 angles:

sin (u+v)= sin u*cos v + sin v*cos u

Let's put arcsin(5/13)=u and arccos(3/5)=v

sin (u+v)=sin (arcsin(5/13) + arccos(3/5))

sin u=sin (arcsin(5/13))=5/13

cos v= cos (arccos(3/5))=3/5

**sin u*cos v=5/13*3/5=3/13**

sin v=sin (arccos(3/5))=sqrt(1-9/25)=sqrt [(25-9)/25]

sin v=sqrt(16/25)=4/5

cos u= cos (arcsin(5/13))=sqrt(1-25/169)=sqrt(144/169)

cos u=12/13

**sin v*cos u=4/5*12/13**

sin (u+v)=3/13 + 4/5*12/13

sin (u+v)=(3*5 + 4*12)/5*13

sin (u+v)=(15+48)/65

**sin (u+v)=63/65**

**Let's substitute u and v, with the innitial values:**

**sin[arcsin(5/13) + arccos(3/5)] =63/65**

To find the value of sin[arcsin(5/13)+arccos(3/5)]:

Let sin [arc sin (x) ] = t, then taking inverse, we get

arc sinx = arc sint or t = x and cos arcsint = sqrt(1-x^2)

Let sin arc cosy = u, then cosy = sinu or u = pi/2-y.So cos arc cosy = y.

Therefore, given expression ,

sin[arcsin(5/13)+arccos(3/5)] = sin(A+B) = sinAcosB+cosAsinB, where, A = arcsin(5/13) and B = arc cos(3/5)

(5/13)[(3/5)+sqrt[1-(5/13)^2]*sqrt[1-(3/5)^2]^(1/2)

=15/68+ (12/13)(4/5) = (15+48)/65 = 63/65.