How do you verify cos^4x - sin^4x = 1 - 2sin^2x? - Socratic
2018年3月19日 · To prove that #cos^4x-sin^4x=1-2sin^2x#, we'll need the Pythagorean identity and a variation on the Pythagorean identity: #color(white)=>cos^2x+sin^2x=1# #=>cos^2x=1-sin^2x# I'll start with the left-hand side and manipulate it until it looks like the right-hand side using these two identities: #LHS=cos^4x-sin^4x#
sin^4x-cos^4x=1-2cos^2x prove it? - Socratic
2018年4月27日 · We will use Pythagoras' identity: #sin^2 x + cos^2 x = 1# from which we can deduce: #sin^2 x = 1 - cos^2 x# Also note that the difference of squares identity can be written:
What is the period and amplitude and frequency for y = cos 4x ...
2015年6月4日 · Amplitude: (-1, 1) since cos 4x varies between -1 and +1. Answer link. Related questions.
What is the period and the fundamental period of #y(x) = sin(2x)
Y(x) is a sum of two trignometric functions. The period of sin 2x would be (2pi)/2 that is pi or 180 degrees. Period of cos4x would be (2pi)/4 that is pi/2,or 90 degrees. Find the LCM of 180 and 90. That would be 180. Hence the period of the given function would be pi
How do you find the antiderivative of #1−cos(4x)#? - Socratic
2016年11月1日 · Say that, as an example but not related to the problem: y=1/k*sin(kx) This would mean that: ky=sin(kx) And ultimately that: k*(dy)/(dx)=k*cos(kx) (dy)/(dx)=cos(kx) Because of the chain rule. Now, if this is the case, when solving the problem : int1-cos(4x)dx =x-1/4sin(4x)+C
What are the first 3 nonzero terms in the Taylor series ... - Socratic
What are the first 3 nonzero terms in the Taylor series expansion about x = 0 for the function #f(x)=cos(4x)#?
How do you graph #y=cos(4x)#? - Socratic
2018年6月7日 · As below. y = cos (4x) Standard form is y = A cos (Bx - C) + D A = 1, B = 4, C = D = 0 Amplitude = |A| = 1 "Period " = (2pi) / |B| = (2pi) / 4 = pi/2 "phase Shift ...
How do you change 2 \sin 7x \cos 4x into a sum? - Socratic
2018年5月23日 · sin 11x + sin 3x f(x) = 2cos 4x. sin 7x Note that sin 7x = cos (pi/2 - 7x) (complementary arcs) f(x) = 2 cos 4x.cos (pi/2 - 7x) Reminder of trig identity: 2cos a.cos b = cos (a - b) + cos (a + b) In this case: a - b = 4x - (pi/2 - 7x) = 11x - pi/2 a + b = 4x + pi/2 - 7x = pi/2 - 3x f(x) = cos (11x - pi2) + cos (pi/2 - 3x) = cos (pi/2 - 11x ...
If #sinx-cosx=1/3#, find #cos(4x)#? - Socratic
2017年12月6日 · cos 4x = - 0.58 x = 58^@63; and x = 211^@37 Use trig identity: sin x - cos x = sqrt2sin (x - pi/4) In this case sqrt2sin (x - pi/4) = 1/3 sin (x - pi/4) = 1/(3sqrt2) = sqrt2/6 Calculator and unit circle give --> a. x - 45 = 13.63 --> x = 45 + 13.63 = 58^@63 4x = 234^@52 --> cos 4x = cos 234^@52 = - 0.58 b. x - 45 = 180 - 13.63 = 166.37 x = 166.37 + 45 = 211^@37 4x = 845.48 = 845.48 - 720 = 125 ...
How do you verify (2cos^2x-1)^2/(cos^4x-sin^4x)=1-2sin^2x
2018年4月30日 · How do you verify #(2cos^2x-1)^2/(cos^4x-sin^4x)=1-2sin^2x#? Trigonometry. 1 Answer