I know that i should multiply something to it for it to become 2sinAcosA for me to be able to solve it. Unfortunately, if i multiply2/2 to it will just become 2/16 that will just bounce back to 1/8...

I know that i should multiply something to it for it to become 2sinAcosA for me to be able to solve it. Unfortunately, if i multiply2/2 to it will just become 2/16 that will just bounce back to 1/8...

I chose to manipulate the left hand side. So i got 1+cos2x-cos2x-cos4x when i simplify it will be 1-cos4x i will break it into pieces so (1-cos2x)(1+cos2x) i can change the first term...

If there is no solution, enter NO SOLUTION.) 2 -2 sin t = 2 sqrt(3)cos t My solutions were pi/2, 7pi/6, and 11pi/6.

Prove 1 - sin2x / cosx = sin2x / 2sinx I tried do both LHS and RHS. Got stuck after a step or two.

Please help, I have no idea! Solve the following trigonometric equation sin^2x - cos^2x = 1/4.

Prove a^2(1+cos)=1-cos if 1-a^2)sin - 2acos =0

(secθ-tanθ)^2=2secθtanθ

I can find x = 0, and also can show that cos x = (1/4)*(-1+/-sqrt(5)), but I don't know how I should have known that the latter cosines of x correspond to angles involving fifths of pi (radians)...

(show work) where 0≤x<2π

Explain how you could determine the exact value of (sec 7pi/6) if you know the value of (sin 11pi/6). Prove with diagrams. I've no idea how to do this question that's worth 10 marks...

This is for my trig class, and the directions say to verify the identity. Thanks so much.

Complete the identity: csc(t)(sin(t) + cos(t)) = ?

tan3x - tan2x- tanx = tan3xtan2xtanx

This is for my trig class. Thanks for the help!

This is for my trig class. Thanks for the help!

1) given the double angle identity for cosine: cos2θ=cos2θ-sin2θ Prove: cos3θ = cos3θ - 3cosθsin2θ and: cos4θ = cos4θ - 6cos2θsin2θ + sin2θ 2)...

2sin^2x - 1÷ sinxcosx = tanx-cotx

cosx = (1- tan^2 x/2) / 1 + tan^2 x/2) Completely stuck.

sin(X)sec(x)=1-(cos^2)x/sin(X)cos(X)

Prove sec2x = 1 + tan2x I did R.H.S = 1 + sin2x / cos2x = 1 + (1 - cos2x) / (1 - sin2x) = 1 + cot2x = ??? I think I'm doing it right,...

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