2016年6月12日 · 1/4sin(2x)+1/2x+C We will use the cosine double-angle identity in order to rewrite cos^2x. (Note that cos^2x=(cosx)^2, they are different ways of writing the same thing.) cos(2x)=2cos^2x-1 This can be solved for cos^2x: cos^2x=(cos(2x)+1)/2 Thus, intcos^2xdx=int(cos(2x)+1)/2dx Split up the integral: =1/2intcos(2x)dx+1/2intdx The second integral is the "perfect integral:" intdx=x+C. =1/2intcos ...

2016年7月11日 · -sin2x >Differentiate using the color(blue)"chain rule" color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(f(g(x)))=f'(g(x))g'(x))color(white)(a/a)|))) .....

2016年7月2日 · How do you differentiate #cos(x^2)#? Calculus Differentiating Trigonometric Functions Differentiating sin ...

2016年10月21日 · #cos^2x=cosx# Let's subtract #cosx# from both sides: #cos^2x-cosx=0# #cosx(cosx-1)=0# Which gives us: #cosx=0, 1# When #cosx=0#, we get: #x=pi/2+npi# When #cosx=1# we get: #x=2pi+n2pi# For both solutions, n is an integer.

2018年5月16日 · 2cos x - cos 2x = -2cos^2x + 2 cos x - 1 # They're very different and your example is not correct. 2 cos x is twice the cosine of angle x. It will be between -2 and 2. cos2x is an abbreviation for cos(2x). It is the cosine of the angle 2x, two times the angle x. The value of cos 2x will be between -1 and 1. Before we think about double angles, let's remember the formula for the cosine of the ...

"The fundamental trigonometric identities" are the basic identities: •The reciprocal identities •The pythagorean identities

Double Angle Identities. #sin2theta=2sin theta cos theta# #cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta#

The best videos and questions to learn about Proving Identities. Get smarter on Socratic.

How do you find the exact value of sinx/2, cosx/2, & tanx/2 given that #sinx= 5/13# is in Quadrant 2? How do you use a half angle formula to find the exact value of sin22.5 degrees? How do you find the exact value of the half angle of #tan 157.5#?

2017年3月25日 · Therefore, the first few derivatives of f(x) are: f(x) = (1+cos(2x))/2 f'(x) = -sin(2x) f''(x) = -2cos(2x) f'''(x) = 4sin(2x) f^((4))(x) = 8cos(2x) Since we are finding a Maclaurin series, plug in 0 to each derivative. f(0) = (1+cos(0))/2 = 1 f'(0) = -sin(0) = 0 f''(0) = -2cos(0) = -2 f'''(0) = 4sin(0) = 0 f^((4))(0) = 8cos(0) = 8 It is easy to ...