2016年4月26日 · 2npi, n=0.+-1,+-2,+-3... The principal values of cos^(-1)(1) in [0,2pi] are 0 and 2pi So, the general value = .2npi, n=0.+-1,+-2,+-3..

2016年9月30日 · x can be any integer multiple of 2pi, including 0 The function cos(x) has period 2pi and cos(0) = 1 Hence: cos(2npi) = 1" " for any integer n graph{cos(x) [-10, 10, -5, 5]}

2015年8月24日 · If #cos^(-1)= theta# #rArrcolor(white)("XXXXX")cos(theta) = -1# This means that the adjacent side is equal in magnitude to the hypotenuse but negative. Within the range #[0,2pi]# this is only true at #theta=pi (= 180^@)# For all solutions (unrestricted in range: #color(white)("XXX")cos^-1(-1)=pi+n2picolor(white)("XXX")AAn in ZZ#

2018年6月1日 · see below cos^(-1)0=arccos 0 What is the arc which cosine is zero?: two posibilities 90=pi/2 and 270=3pi/2 This is assuming that cos^(-1) is the inverse of cosine. There is no missunderstanding if use arccos instead of cos^(-1) Because cos^(-1) is also understud as 1/(cos)=sec which is different

2015年6月28日 · In terms of the right triangles used to define trigonometric functions, cos(x) = frac{"adjacent side"}{"hypotenuse"}. When x=0, "adjacent side length" = "hypotenuse length". Therefore, cos(0) = 1. Consider a series of triangles with …

2016年9月7日 · =x cos^(-1) x - sqrt(1-x^2) + C int cos^(-1) x dx we know d/dx ( cos^(-1) x ) = -1/sqrt(1-x^2) so we can try set up an IBP and use that fact = int d/dx (x) cos^(-1) x ...

2016年5月9日 · sin(cos^(-1)(x)) = sqrt(1-x^2) Let's draw a right triangle with an angle of a = cos^(-1)(x). As we know cos(a) = x = x/1 we can label the adjacent leg as x and the hypotenuse as 1. The Pythagorean theorem then allows us to solve for the second leg as sqrt(1-x^2). With this, we can now find sin(cos^(-1)(x)) as the quotient of the opposite leg and the hypotenuse. sin(cos^( …

2017年2月7日 · d/dxcos^(-1)(x) = -1/sqrt(1 -x^2) When tackling the derivative of inverse trig functions. I prefer to rearrange and use Implicit differentiation as I always get the inverse derivatives muddled up, and this way I do not need to remember the inverse derivatives. If you can remember the inverse derivatives then you can use the chain rule. Let y=cos^(-1)(x) <=> …

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2016年8月6日 · #arccos x = 1/2# Trig table of special arcs gives #cos x = 1/2#--> #x = pi/3# Trig unit circle gives another arc x that has the same cos value -->