2017年8月4日 · 4x^3cos(x^4) By the chain rule (sin(x^4))'=(x^4)'sin'(x^4) =4x^3sin'(x^4) by the power rule And since sin'(x)=cos(x) (sin(x^4))'=4x^3cos(x^4)

#f(x) = cos^4x-sin^4x# First let's do some simplification. Notice: #f(x) = (cos^2x+sin^2x)(cos^2x-sin^2x)# Since # (cos^2x+sin^2x) = 1 -> f(x) = (cos^2x-sin^2x)#

2018年2月6日 · 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 …

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:

2018年3月12日 · It is a perfect square: sin^4(x)-2sin^2(x)+1= (sin^2(x)-1)^2. What are the units used for the ideal gas law

2018年4月5日 · #sin ^4 x - cos ^4 x = cos 3x # #(sin ^2 x + cos ^2 x)(sin ^2 x - cos ^2 x ) = cos 3x # #-cos 2x = cos 3x # #cos (180^circ - 2x) = cos 3x # I've been on Socratic for a couple of …

2016年6月5日 · \frac{1}{16}\(x-\frac{1}{4}\sin \(4x\))+\frac{1}{6}\sin ^3\(x\)\cos ^3\(x)+C \int \sin^2\(x\)cos^4\(x\)dx Applying integral reduction, int sin^2(x) cos^n (x) dx ...

2016年9月18日 · See below. Solving for y sin^4(x)/x + cos^4(x)/y = 1/(x + y) we obtain after some simplifications y = x (cos(x)/sin(x))^2 = x cot^2(x) Now, substituting back in (sin x)^12 /x^5 + …

2015年12月5日 · #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

2018年5月20日 · Use trig identity: #sin^2 x = 1/(1 + cot^2 x)# In this case tan x = 4 --> #cot x = 1/4# #sin^2 x = 1/(1 + 1/16) = 16/17#