need help with this

Question

find the exact value of cos(17π/6)=x , and sin(300 degrees)=y

Answer 1

Steve S. has a much nicer and simpler solution to the first problem and the second problem also.

However, here is another solution to the first one.

cos(17pi/6)=cos(17*30)=cos(510)

510-360=150

cos(150)=cos(180-30)

cos(180-30)=cos(180)cos(30)+sin(180)sin(30

                  =(-1)(√3/2)+(0)(1/2)

cos(150)=-√3/2 so cos(510)=-√3/2 which is exact and not approximate like my first answer

 

 

 

 

sin(300º)=-0.86602540378

sin(300º)=sin(270º+30º)

sin(270+30)=sin(270)cos(30)+cos(270)sin(30)

                  =(-1)(√3/2)+(0)(1/2)

                  =(-1)(0.86602540378)

                  =-0.86602540378

sine is negative in the fourth quadrant

 

Answer 2

cos(17π/6)=x , and sin(300 degrees)=y

 

17 pi/6 = 12 pi/6 + 6 pi/6 - pi/6 = 2 pi + pi - pi/6.

 

So the angle lies in the 2nd Quadrant where the cosine is negative (the x coordinates of points there are negative and x = r cosθ). The reference angle is pi/6 = 30° and we know its cosine from the Special 30°-60°-90° Triangle to be √(3)/2.

 

Putting it all together: cos(17π/6) = - cos(π/6) = -√(3)/2.

 

300° = 360° - 60° so the angle lies in the 4th Quadrant where sine is negative (the y coordinates of points there are negative and y = r sinθ). The reference angle is 60° and we know its sine from the Special 30°-60°-90° Triangle to be √(3)/2.

 

Putting it all together: sin(300°) = - sin(60°) = -√(3)/2.