How do you solve #sin(105^@)=sin(60^@+45^@)#?

1 Answer
Douglas K.
Dec 7, 2017

Use the identity #sin(A+B) = sin(A)cos(B)+cos(A)sin(B)#.
The values for the sine and cosine of #60^@# and #45^@# are well known; use them to simplify the result.

Explanation:

Given:

#sin(105^@)=sin(60^@+45^@)#

Use the identity #sin(A+B) = sin(A)cos(B)+cos(A)sin(B)# where #A=60^@# and #B=45^@#:

#sin(105^@)=sin(60^@)cos(45^@)+cos(60^@)sin(45^@)#

The sine and cosine of #45^@# are both #sqrt2/2#:

#sin(105^@)=sin(60^@)(sqrt2/2)+cos(60^@)(sqrt2/2)#

The sine of #60^@# is #sqrt3/2# and the cosine is #1/2#

#sin(105^@)=sqrt3/2(sqrt2/2)+1/2(sqrt2/2)#

#sin(105^@)=(sqrt6+ sqrt2)/4#

Related questions
Impact of this question
3258 views around the world
You can reuse this answer
Creative Commons License