How do you find the exact functional value Sin 165 using the cosine sum or difference identity?

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Answer 1 · 2018-03-28T16:18:47

#sin165^0=-((sqrt6-sqrt2)/4)#

We know that,

#color(red)((1)sin(pi/2+theta)=-costheta#

#color(red)((2)cos(A+B)=cosAcosB-sinAsinB#

Here,

#sin165^0=sin(90^0+75^0)....tocolor(red)(.Apply (1)#

#=-cos75^0#

#=-cos(45^0+30^0).....tocolor(red)(Apply(2)#

#=-[cos45^0cos30^0-sin45^0sin30^0]#

#=-[1/sqrt2xxsqrt3/2-1/sqrt2xx1/2]#

#=-[(sqrt3-1)/(2sqrt2)]#

#=-[(sqrt3-1)/(2sqrt2)xxsqrt2/sqrt2]#

#=-((sqrt6-sqrt2)/4)#