How do you find the value of #sin(112 1/2)# using the double or half angle formula?

Question

4386 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-22T13:53:41

#sin(112 1/2)^o=sqrt(2+sqrt2)/2#

As #sin(90^o+theta)=costheta#

#sin(90^o +(22 1/2)^o)=cos(22 1/2)^o#

Now as #2cos^2A-1=cos2A#, if #A=(22 1/2)^o#, #A=45^o#

and #2cos^2(22 1/2)^o-1=cos45^o=1/sqrt2#

or #2cos^2(22 1/2)^o=1+1/sqrt2=(sqrt2+1)/sqrt2#

and #cos^2(22 1/2)^o=(sqrt2+1)/(2sqrt2)=(2+sqrt2)/4#

#cos(22 1/2)^o=sqrt(2+sqrt2)/2#

Hence #sin(112 1/2)^o=sqrt(2+sqrt2)/2#