2016年6月9日 · How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle

2018年3月7日 · #cos(pi/4)=sqrt(2)/2#, refer to the explanation below for how to find the exact value without a calculator. Explanation: It is possible to find the exact value of #cos(pi/4)# by constructing a right triangle with one angle set to #pi/4# radians.

2017年12月19日 · #cos((2pi)/9)*cos((4pi)/9)*cos((8pi)/9)# = #1/2cos((4pi)/9)[2cos((8pi)/9)cos((2pi)/9)]# = #1/2cos((4pi)/9)[cos((8pi)/9+(2pi)/9)+cos((8pi)/9-(2pi)/9)]#

2018年7月3日 · How do you evaluate #cos ((4pi)/3)#? Trigonometry Right Triangles Trigonometric Functions of Any Angle. 1 ...

2016年5月2日 · Why is cos(4pi/3) = - cos(pi/3)? Trigonometry Right Triangles Relating Trigonometric Functions. 1 Answer

2016年8月15日 · Trig table of special arcs, and unit circle --> #cos ((-4pi)/3) = cos ((4pi)/3) = cos (pi/3 + pi) = - cos (pi/3) = -1/2#

2016年10月3日 · #cos(pi/7)cos((4pi)/7) cos((5pi)/7)# #=4/(8sin(pi/7))xx(2sin(pi/7)cos(pi/7))cos((4pi)/7)cos(pi-(2pi)/7)# #=-2/(8sin(pi/7))xx(2sin((2pi)/7)cos((2pi)/7))cos((4pi)/7)#

2016年11月20日 · De Moivre's formula tells us that: #(cos theta + i sin theta)^n = cos n theta + i sin n theta# So we find: #(2(cos(pi/3)+i sin(pi/3)))^4 = 2^4(cos((4pi)/3) + i sin ...

2016年3月3日 · Multiply both sides with #sin(pi/7)# hence we have that. #sin(π/7)cos(2π/7)+sin(π/7)cos(4π/7)+sin(pi/7)*cos(6pi/7)=-1/2*sin(π/7)#

2016年7月3日 · -1/2 Use the trig identity: cos 2a = 2cos^2 a - 1 cos ((4pi)/3) = 2cos^2 ((2pi)/3) - 1 Trig table gives: cos ((2pi)/3) = -1/2.