How do you simplify #sin[cos^-1( - sqrt5 / 5 ) + tan^-1 ( - 1 / 3) ]#?

1 Answer
May 26, 2016

#=+-0.7 sqrt 2 and +-1/sqrt 2=+-0.98995 and +-0.70711#, nearly.

Explanation:

Let #a = cos^(-1)(-sqrt 5/5)#. Then #cos a = -sqrt 5/5=-1/sqrt 5<0#. so, a is in the 2nd quadrant or in the 4th. Accordingly, #sin a = +-2/sqrt 5#.

Let #b = tan^(-1)(-1/3)#. Then #tan b = -1/3<0#. So, b is in the 2nd quadrant or in the 4th. Accordingly, #sin b = +-1/sqrt 10 and cos b = 3/sqrt 10#. Also, sin b and cos b have opposite signs.

Now, the given expression is

sin ( a + b ) = sin a cos b + cos a sin b

#=(+-2/sqrt 5)(+-3/sqrt 10)-(-1/sqrt 5)(+-1/sqrt 10)#

#=+-6/sqrt 50# #-# or + #1/sqrt 50#

#=(sqrt 2/10)(+-6+-1)#

#=+-0.7 sqrt 2 and +-1/ sqrt 2#

In each case, the angles a and b can be obtained separately. .