The period is 2*pi radians.
y=sinx y=cosxsinx=cosx=>sinx/cosx=1=>tanx=1=>x=45oie.. y=sin45=cos45y=1/(square root of 2)
2
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
cosx√y=5 Right away you can tell that there'll be a domain error because if you were to take cosine inverse five equals X root y then there would be an error
From the Pythagorean identity, sin2x = 1-cos2x. LHS = 1/(sinx cosx) - cosx/sinx LHS = 1/(sinx cosx) - (cosx/sinx)(cosx/cosx) LHS = 1/(sinx cosx) - cos2x/(sinx cosx) LHS = (1- cos2x)/(sinx cosx) LHS = sin2x /(sinx cosx) [from Pythagorean identity] LHS = sin2x /(sinx cosx) LHS = sinx/cosx LHS = tanx [by definition] RHS = tanx LHS = RHS and so the identity is proven. Q.E.D.
The same as the period of y = sin x. This period is equal to (2 x pi).
Remember SecX = 1/CosX Substitute SinX X 1 /CosX = SinX / CosX = TanX
the period is 2pi. period is 2pi/b and the formula is y=AsinBx.
x = 3pi/4
pi radians.
The period is the length of x over which the equation repeats itself. In this case, y=sin x delivers y=0 at x=0 at a gradient of 1. y next equals 0 when x equals pi, but at this point the gradient is minus 1. y next equals 0 when x equals 2pi, and at this point the gradient is 1 again. Therefore the period of y=sinx is 2pi.
2sinxcosx-cosx=0 Factored : cosx(2sinx-1)=0 2 solutions: cosx=0 or sinx=.5 For cosx=0, x=90 or 270 degrees For sinx=.5, x=30 degrees x = {30, 90, 270}