How do you express #sin(pi/12) * cos(( 13 pi)/12 ) # without products of trigonometric functions?

1 Answer
Jul 3, 2018

#color(red)(-sin (pi/6)= -1/2#

Explanation:

![https://study.com/academy/lesson/http://product-to-sum-identities-uses-applications.html](https://useruploads.socratic.org/2xBiCxzSmmCr1qNXSs2e_trigonometric%20identities.png)

#sin (pi/12) * cos ((13pi)/12)#

#sin x cos y = (1/2)[sin(x+y) + sin (x-y])#

#=> (1/2)[sin (pi/12 + (13pi)/12) + sin (pi/12 - (13pi)/12)]#

#=> [sin ((14pi)/12) + sin (-pi)]/2#

#=> [sin (pi + pi/6) - sin (pi)]/2#

#color(red)(-sin (pi/6)= -1/2#