Mar 19, 2018 · 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#

Jun 7, 2018 · 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 ...

Dec 19, 2017 · #cos(f_1) df_1 = cos(4x) 4 dx = 4 cos(4x) dx# Now, the tiny nudge is no longer in terms of #df_1#, but it is now in terms of #dx#: #df_2 = 4 cos(4x) dx rarr (df_2)/(dx) = 4 cos(4x)# And so is the derivative! What we have left to do is #f_3(x)#, whose tiny nudge we'll first find in terms of #df_2#:

Jun 4, 2015 · Amplitude: (-1, 1) since cos 4x varies between -1 and +1. Answer link. Related questions.

May 14, 2018 · How your verify #1/8[3+4cos2x+cos4x]=cos^ 4x#? Trigonometry. 1 Answer Abhishek K. May 14, 2018 #RHS=cos^4x

Nov 1, 2016 · 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

Dec 5, 2015 · #cos^4 x - sin^2 x = (cos ^2 x - sin^2 x( (cos^2 x + sin^2 x) = # #= (cos@ x - sin^2 x).# because (cos^2 x + sin^2 x) = 1

What are the first 3 nonzero terms in the Taylor series expansion about x = 0 for the function #f(x)=cos(4x)#?

May 3, 2018 · pi^2/4. We will use these Results : (R_1) : int_0^af(x)dx=int_0^af(a-x)dx. (R_2)(i) : int_0^(2a)f(x)dx=0, if f(2a-x)=-f(x), and, (R_2)(ii) : int_0^(2a)f(x)dx=2int_0 ...

Feb 6, 2018 · cos^5x This type of problem is truly not that bad once you recognize that it involves a little algebra! First, I'll rewrite the given expression to make the following steps easier to understand. We know that sin^2x is just a simpler way to write (sin x)^2. Similarly, sin^4x = (sin x)^4. We can now rewrite the original expression. (sin^4 x - 2 sin^2 x +1) cos x =[ (sin x)^4 - …