How do you evaluate #sin((5pi)/9)cos((7pi)/18)-cos((5pi)/9)sin((7pi)/18)#?

1 Answer
Apr 16, 2016

#1/2#

Explanation:

This equation can be solved using some knowledge about some trigonometric identities. In this case, the expansion of #sin(A-B)# should be known:

#sin(A-B)=sinAcosB-cosAsinB#

You'll notice that this looks awfully similar to the equation in the question. Using the knowledge, we can solve it:
#sin((5pi)/9)cos((7pi)/18)-cos((5pi)/9)sin((7pi)/18)#
#=sin((5pi)/9-(7pi)/18)#
#=sin((10pi)/18-(7pi)/18)#
#=sin((3pi)/18)#
#=sin((pi)/6)#, and that has exact value of #1/2#