Differentiate A) y = -Cos2x

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Solve the equation of the interval (0, 2pi)cosx=sinx I squared both sides to get :cos²x=sin²x Then using tri indentites I came
1. 4 answers
2. asked by Brian
3. 780 views
Solve identity,(1-sin2x)/cos2x = cos2x/(1+sin2x) I tried starting from the right side, RS: =(cos²x-sin²x)/(1+2sinxcosx)
1. 3 answers
2. asked by sh
3. 734 views
1. differentiate cos(3/x)2. differentiate sin(4/x) 3. differentiate 3/{sin(3x+pi)} 4. differentiate pxsin(q/x)where p and q are
1. 0 answers
2. asked by I'm stumped
3. 767 views
Find all solutions to the equation in the interval [0, 2pi)cos4x-cos2x=0 So,this is what i've done so far: cos4x-cos2x=0
1. 1 answer
2. asked by Maria
3. 1,623 views
Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π).8. cos2x=cosx 10. 2cos^2x+cosx=cos2x Solve
1. 5 answers
2. asked by Anonymous
3. 815 views
DifferentiateA) y = -Cos2x B) y = Sin2tetta - 2Cos2tetta C) f(tetta) = negative pie*Sin(2tetta - pie) Tetta means that zero with
1. 3 answers
2. asked by Julie
3. 422 views
find all solutions of the equation 2sin x cos2x-cos2x=0 over the interval 0<x<=pi
1. 4 answers
2. asked by owo
3. 2,200 views
Prove that cos3x/cosx-cos6x/cos2x=2(cos2x-cos4x)
1. 2 answers
2. asked by Joel
3. 1,150 views
Where do I start to prove this identity:sinx/cosx= 1-cos2x/sin2x please help!! Hint: Fractions are evil. Get rid of them. Well,
1. 0 answers
2. asked by maria
3. 804 views
Use the Pythagorean identity to show that the double angle formula for cosine can be written asa) cos2x = 1 - 2sin^2x b) cos2x =
1. 1 answer
2. asked by Anonymous
3. 1,641 views