Cos 16pi/9 cos 10pi/9 + sin 16pi/9 sin 10pi/9

Question

I don't know how to solve

Answer 1

cosAcosB+sinAsinB=cos(A-B)

A=16pi/9

B=10pi/9

using the identity...

cos(A-B)=cos(16pi/9-10pi/9)

cos(16pi/9-10pi/9)=cos(6pi/9)

cos(6pi/9)=cos(2pi/3)

cos(2pi/3)=cos(2*60)=cos(120)

cos(120º)=-0.5 using arcos on your calculator or a table or look on the unit circle: cos(2pi/3)=-1/2 or -0.5

Answer 2

cosA cosB + sinA sinB = cos(A-B)

 

Using this formula, we get:

cos(16pi/9) cos(10pi/9) + sin(16pi/9) sin(10pi/9) = cos(16pi/9 - 10pi/9) = cos(6pi/9) = cos(2pi/3) = cos(120 degrees) = -1/2.

 

(Remember that the angles are given here in radians)