Let a = cos^(-1)(-sqrt 5/5)a=cos−1(−√55). Then cos a = -sqrt 5/5=-1/sqrt 5<0cosa=−√55=−1√5<0. so, a is in the 2nd quadrant or in the 4th. Accordingly, sin a = +-2/sqrt 5sina=±2√5.
Let b = tan^(-1)(-1/3)b=tan−1(−13). Then tan b = -1/3<0tanb=−13<0. So, b is in the 2nd quadrant or in the 4th. Accordingly, sin b = +-1/sqrt 10 and cos b = 3/sqrt 10sinb=±1√10andcosb=3√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)=(±2√5)(±3√10)−(−1√5)(±1√10)
=+-6/sqrt 50=±6√50 -− or + 1/sqrt 501√50
=(sqrt 2/10)(+-6+-1)=(√210)(±6±1)
=+-0.7 sqrt 2 and +-1/ sqrt 2=±0.7√2and±1√2
In each case, the angles a and b can be obtained separately. .
