How do you express #sin(pi/12) * cos(( pi)/2 ) # without products of trigonometric functions?

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Answer 1 · 2016-06-14T15:33:21

#color(blue)(sin (pi/12)*cos (pi/2)=1/2sin ((7pi)/12)-1/2sin ((5pi)/12))#

Let us use the sum and difference formulas

#sin (x+y)=sin x*cos y +cos x* sin y#

#sin (x-y)=sin x*cos y -cos x* sin y#

After addition of equal values, we have

#sin (x+y)+sin (x-y)=2*sin x*cos y#

and we have the formula

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

We can now use the given

#sin (pi/12)*cos (pi/2)=1/2sin (pi/12+pi/2)+1/2sin (pi/12-pi/2)#

#sin (pi/12)*cos (pi/2)=1/2sin ((7pi)/12)+1/2sin ((-5pi)/12)#

#sin (pi/12)*cos (pi/2)=1/2sin ((7pi)/12)-1/2sin ((5pi)/12)#

God bless....I hope the explanation is useful.