The curve with equation y=(sin 1/2x)/2+cos 1/2x is defined for 0<x<2pi.
Show that dy/dx=(1/2 + cos 1/2x)/(2+cos 1/2x)^2
I've used the quotient rule and ended up with dy/dx=(2+cos 1/2x((cos 1/2x)/2)-sin 1/2x((-sin 1/2x)/2))/(2+cos 1/2x)^2 and not sure where to go from there. Would really appreciate some help please, thanks
The curve with equation y=(sin 1/2x)/2+cos 1/2x is defined for 0<x<2pi.
Show that dy/dx=(1/2 + cos 1/2x)/(2+cos 1/2x)^2
I've used the quotient rule and ended up with dy/dx=(2+cos 1/2x((cos 1/2x)/2)-sin 1/2x((-sin 1/2x)/2))/(2+cos 1/2x)^2 and not sure where to go from there. Would really appreciate some help please, thanks
Write out the numerator again, then multiply out the sin and cos and apply two identities to the resulting terms.
but I have sin^2 (x/2) and cos^2 (x/2) instead so what do I do?
You can apply the normal identity. All that sin2α+cos2α=1 means is that the sum of the squares of the sine and cosine of a given angle α is 1. x/2 is still an angle so the identity holds for that also.
Ok thanks guys I got to the answer. So the next part is: find the value of x at which the gradient of the curve is 1/12, giving your answer to 3sf. So 1/12=(1/2 + cos (x/2)/(2 + cos (x/2))^2 , but I keep getting stuck on rearranging. Any tips?