How do you evaluate #Sin((5pi)/12) cos(pi/4) - cos((5pi)/12) sin(pi/4)#?

2 Answers
Shwetank Mauria
Mar 3, 2016

#0.5#

Explanation:

To evaluate #sin((5pi)/12)cos(pi/4)−cos((5pi)/12)sin(pi/4)#, we can use the identity

#sin(x-y)=sinxcosy-cosxsiny#

Hence #sin((5pi)/12)cos(pi/4)−cos((5pi)/12)sin(pi/4)#

= #sin((5pi)/12-pi/4) = sin((5pi)/12-(3pi)/12) # or

= #sin((2pi)/12)=sin(pi/6)=0.5#

Nghi N.
Mar 3, 2016

#sin (pi/6) = 1/2#

Explanation:

Trig identity: sin (a - b) = sin a.cos b - sin b.cos a.
Therefor, the expression is reduced to:
#sin ((5pi)/12 - sin (pi/4)) = sin ((2pi)/12) = sin (pi/6) = 1/2#

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