Alright I have a function y = sin2x-2sinx and I have to find x-coordinates to all the points on the graph that have a tangent line that is horizontal, or zero. So I'm guessing chain rule along with the derivatives of sin which I can do but how to find the points where it is zero? Find the derivative of this function then plug in zero? Thanks for any help.
y = sin2x - 2sinx
=> y' = 2cos2x - 2cosx
for horizontal slope, set y' = 0
=> 2cos2x - 2cosx = 0
=> cos2x - cosx = 0
=> cos^2x - sin^2x - cosx = 0
=> cos^2x - (1 - cos^2x) - cosx = 0
=> 2cos^2x - cosx - 1 = 0
let y = cosx
=> 2y^2 - y - 1 = 0
=> (2y + 1)(y - 1) = 0
=> y = -1/2 or y = 1
=> cosx = -1/2 or cosx = 1
so x = cos^-1(-1/2) or x = cos^-1(1) .............cos^-1 is cosine inverse. plug this into your calculator to find a value and remember you have to add + kpi for all solutions
Ok so I kind of followed your work there. Kinda fell off at => cos2x - cosx = 0
=> cos^2x - sin^2x - cosx = 0. But then at the end if I plug in the values you got and trace them on the original graph of y = sin2x - 2sinx and the points don't match up with where a logical horizontal tangent appears on the graph. I'm probably just doing something wrong but what I can guess is horizontal tangents at 1.11111 and -1.111111 then at 4 other points but inverse cosine of 1 is 0 which on the graph isn't a max or min value. I'm pretty lost I think.
cos2x = cos^2x - sin^2x is a trig identity.
it turns out that the values gained for the cos^-1(1) solutions are inflection points. you can check this with the second derivative. it's not wrong, the derivative is zero at inflection points as well, they are technically critical points
No I posted the original problem correctly. I'm using my TI-86 to graph the function and by the looks of it I should have 6 horizontal tangents. And when I bring up the menu on the calculator to trace and I enter cos^-1 (1) its 0 and 0 is on a curve not a min or max where the tangent line would be horiztonal. And my calculator is in radians mode so I know thats not the problem.