How do you use the sum and difference identities to find the exact value of #cos 15^@#?

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Answer 1 · 2014-10-24T04:22:09

The special triangles, 30-60-90 and 45-45-90, allow us to evaluate sin and cos.

We leverage that information to evaluate #cos(15)#.

#cos(a-b)=cos(a)cos(b)+sin(a)sin(b)#

#cos(60-45)=cos(60)cos(45)+sin(60)sin(45)#

#cos(15)=cos(60)cos(45)+sin(60)sin(45)#

#cos(15)=1/2*1/sqrt(2)+sqrt(3)/2*1/sqrt(2)#

#cos(15)=1/(2sqrt(2))+sqrt(3)/(2sqrt(2))#

#cos(15)=(1+sqrt(3))/(2sqrt(2))#

#cos(15)=(1+sqrt(3))/(2sqrt(2))*sqrt(2)/sqrt(2)#

#cos(15)=(sqrt(2)+sqrt(6))/4#

#cos(15)=0.9659258263#

Verify your results using a calculator. Make sure the calculator is in Degree Mode.

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