How do you use the angle sum or difference identity to find the exact value of #cos(pi/3+pi/6)#?

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Answer 1 · 2018-06-26T06:09:31

#cos(A+B)=cos(A)cos(B)-sin(A)sin(B)#

Therefore #cos(pi/3+pi/6)# turns into

#cos(pi/3)cos(pi/6)-sin(pi/3)(sin(pi/6)#

#1/2*sqrt3/2-sqrt3/2*1/2 rarr 0#

This makes sense as #cos(pi/3+pi/6) rarr cos(pi/2) rarr 0#

Answer 2 · 2018-06-26T06:10:55

#cos (pi/3 + pi/6) = 0#

#cos (A + B) = cos A cos B - sin A sin B#

#hat A = pi/3, hat B = pi/6#

#cos A = cos (pi/3) = 1/2, sin A = sin (pi/3) = sqrt3 / 2#

#cos B = cos (pi/6) = sqrt3 / 2, sin B = sin (pi/6) = 1 / 2#

#cos (pi/3 + pi/6) = ((1/2) * sqrt3/2) - (sqrt3/2 * (1/2)) = 0#

Answer 3 · 2018-06-26T06:12:54

#cos (pi/3 + pi/6) = cos (pi/2) = 0#

#cos (pi/3 + pi/6)#

#=> cos ((2pi + pi) / 6 ) = cos (pi/2) = cos (90^@) = 0#