Mark M. answered • 09/30/18
Tutor
4.9
(954)
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
sin(θ/2) = ±√([1-cosθ] / 2) and cos(θ/2) = ±√[(1+cosθ)/2]
So, sin(5x) = sin((10x)/2) = ±√[(1-cos(10x))/2]
Therefore, [sin(5x)]4 = [(1-cos(10x))/2]2 = [1 - 2cos(10x) + cos2(10x] / 4
= 1/4 - (1/2)cos(10x) + (1/4)[(1 + cos(20x))/2]
= 1/4 - (1/2)cos(10x) + 1/8 + (1/8)cos(20x)
= 3/8 - (1/2)cos(10x) + (1/8)cos(20x)
Mark M.
tutor
I did use the half angle formula for sine in line 2, and then squared the square root in line 3. Also, cos^2(10x)/4 = [cos^2(20x/2)]/4 = what I got in line 4 by squaring the square root in the half angle formula for cosine.
Report
10/18/21
Kyle H.
3 years late, but you should have clarified that (1 + cos(20x))/2 from the 4th and 5th lines came from the reduction formula for cos: cos^2θ = (1 + cos(2θ)) / 2 It doesn't make any sense otherwise how you get from (cos^2(10x) / 4 to (1 + cos(20x)) / 2. Furthermore, you didn't need to write the half-angle formulas for sin and cos at the beginning. You don't use the square root anywhere else in the solution so that will just throw viewers off. The same steps can be reached with the reduction formula for sin: sin^2θ = (1 - cos(2θ)) / 2 = (1 - cos(2 * 5x)) / 210/17/21