2015年10月18日 · Express sin (x/2) in terms of cos x. Ans: sin (x /2) = sqrt((1 - cos x)/2) By applying the trig identity: cos 2a = 1 - 2sin^2 a, we get: cos x = 1 - 2sin^2 (x/2) 2sin^2 (x/2) = 1 - cos x sin^2 (x/2) = (1 - cos x)/2 sin (x/2) = +- sqrt((1 - cos x)/2)

2005年11月3日 · So, sin^2(x) = (sinx)^2 = sinx*sinx? Nov 3, 2005 #4 masudr. 931 0. Yes! I have to extend the message. Nov ...

2017年3月7日 · Calculus Differentiating Trigonometric Functions Differentiating sin(x) from First Principles. 1 Answer . Alan N.

2017年3月9日 · How do you verify the identity #sin(x/2)cos(x/2)=sinx/2#? Trigonometry Trigonometric Identities and Equations Half-Angle Identities. 1 Answer

2017年11月12日 · x^2 - x^6/(3!) + x^10/(5!) - .... sum_(n=0 )^oo x^(4n+2)/((2n+1)!) * (-1)^n First we must find the series for sin(x) let f(x) = sin(x) f(0) = sin(0) = 0 f'(0) = cos(0 ...

2015年6月11日 · How do you find the taylor series of #sin (x^2)#? Calculus Power Series Constructing a Taylor Series. 1 Answer

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

2014年9月8日 · Answer 2sin(x)cos(x) Explanation You would use the chain rule to solve this. To do that, you'll have to determine what the "outer" function is and what the "inner" function composed in the outer function is. In this case, sin(x) is the inner function that is composed as part of the sin^2(x). To look at it another way, let's denote u=sin(x) so that u^2=sin^2(x). Do you …

2018年6月22日 · Note that we can write this more compactly as #csc^2x# as cosec #x# is #1/sinx#. But to take the derivative this form in terms of sin is more useful. Use the quotient rule for the overall fraction and the chain rule to differentiate #sin^2x# .

2016年10月30日 · #e^y=(sinx)^2# Use implicit differentiation on the left hand side of the equation and the chain rule on the right hand side of the equation: #e^y*(dy)/(dx)=2sinx*cosx#