Find the value of #sin(cos^(-1)(1/2)+tan^(-1)(1))#?

1 Answer
Dec 24, 2017

#sin(cos^(-1)(1/2)+tan^(-1)(1))=(sqrt6+sqrt2)/4#

Explanation:

To find #sin(cos^(-1)(1/2)+tan^(-1)(1))#, let us first find #cos^(-1)(1/2)# and #tan^(-1)(1)#.

The range of #cos^(-1)x# is #[0,pi]# and that of #tan^(-1)x# is #[-pi/2,pi/2]#.

As such #cos^(-1)(1/2)=pi/3# and #tan^(-1)(1)=pi/4#

Hence, #sin(pi/3+pi/4)=sin(pi/3)cos(pi/4)+cos(pi/3)sin(pi/4)#

= #sqrt3/2xx1/sqrt2+1/2xx1/sqrt2#

= #(sqrt3+1)/(2sqrt2)#

= #(sqrt6+sqrt2)/4#