How do you find #sin(sin^-1(1/2)+cos^-1(3/5))#?

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Answer 1 · 2016-08-30T16:24:41

#(3+4sqrt 3)/10#.

Let #a = sin^(-1)(1/2) in Q1, for the principal value#. Then,

#sin a = 1/2 and cos a = sqrt(1-sin^2 a) = sqrt(1-1/4) =sqrt 3/2#, for a in

Q1.

Let #b = cos^(-1)(3/5) in Q1, for the principal value#. Then,

#cosb = 3/5 and sin b = sqrt(1-cos^2 a) = sqrt(1-9/25) =4/5#, for b in

Q1.

Now, the given expression is

#sin(a+b)#

#=sina cos b +cos a sin b#

#=(1/2)(3/5)+(sqrt 3/2)(4/5)#

#(3+4sqrt 3)/10#.

Yet, a could be in Q3, wherein #cos a = - sqrt 3/2#, and, similarly, b

could be in Q4, wherein #sin b = -4/5#. Considering this, the

general value is

#+-3+-4sqrt 3)/10#,

when the principal-value convention is relaxed..