How do you express #sin(pi/ 6 ) * cos( ( pi) / 12 ) # without using products of trigonometric functions?

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Answer 1 · 2016-01-29T16:45:12

Either evaluate or use the product to sum identity.

Evaluate
We have:

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

and

#cos(pi/12) = cos(pi/4-pi/6)=cos(pi/4)cos(pi/6)+sin(pi/4)sin(pi/6) = (sqrt6+sqrt2)/4#.

So #sin(pi/6) cos(pi/12) = 1/2 * (sqrt6+sqrt2)/4 = (sqrt6+sqrt2)/8#

Use the product to sum identity.

#sin(pi/6)cos(pi/12) = 1/2 [sin((pi/6)+(pi/12)) + sin((pi/6)-(pi/12))]#

# = 1/2 [ sin(pi/4)+sin(pi/12)]#