2016年10月8日 · This means that sin^(-1)sin(100pi)=100pi, For problems in applications tn which x = a function of time, the principal-value-convention has to be relaxed. Having noted that there were 2.8 K viewers, I add more, to introduce my piecewise-wholesome inverse operators for future computers, for giving the answer as x for any x in ( -oo, oo ).

2016年4月15日 · 1/sqrt(1-x^2) Let y=sin^-1x, so siny=x and -pi/2 <= y <= pi/2 (by the definition of inverse sine). Now differentiate implicitly: cosy dy/dx = 1, so dy/dx = 1/cosy.

2018年6月12日 · 17171 views around the world You can reuse this answer ...

2015年4月15日 · If x is a non-right angle in a right angled triangle then sin(x) is the ratio of the length of the side opposite x with the hypotenuse of the triangle If we restrict our answer to x within [0,2pi] sin(x) = 1 only occurs when x=pi/2

2017年6月22日 · The limit does not exist. To understand why we can't find this limit, consider the following: We can make a new variable h so that h = 1/x. As x -> 0, h -> oo, since 1/0 is undefined. So, we can say that: lim_(x->0)sin(1/x) = lim_(h->oo)sin(h) As h gets bigger, sin(h) keeps fluctuating between -1 and 1. It never tends towards anything, or stops fluctuating at any point. So, we can say that the ...

2014年9月12日 · #lim_(1/xrarr0)sin (1/x)/(1/x)#. With #h=1/x#, this becomes #lim_(hrarr0)sin h/h# which is #1#. Although it is NOT needed, here's the graph of the function: graph{y = x sin(1/x) [-5.55, 5.55, -2.775, 2.774]}

2016年10月21日 · ±sqrt (1-x^2) cos(sin^-1 x) Let, sin^-1x = theta =>sin theta = x =>sin^2theta =x^2 =>1-cos^2theta = x^2 =>cos^2theta = 1-x^2 =>cos theta =± sqrt (1-x^2) =>theta ...

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^(-1)(x)) = sin(a) = sqrt ...

2015年10月17日 · First, let's call #sin(tan^-1(x))=sin(theta)# where the angle #theta=tan^-1(x)#. More specifically, #tan^-1(x)=theta# is the angle when #tan(theta)=x# . We know this from the definition of inverse functions.

2015年4月16日 · Let's say your expression is called E. First, multiply the first fraction by "1-sinx" and the second by "1+sinx" E = (1-sinx)/((1+sinx) * (1-sinx)) + (1+sinx)/((1 ...