I thought it'd be (1 / cos(θ)2) * (1 / tan(θ)) * (cos(θ) / 1) Which would get you cos(θ) / cos(θ)2 * tan(θ) And then coming out to 1 / cos(θ) *...

I thought it'd be (1 / cos(θ)2) * (1 / tan(θ)) * (cos(θ) / 1) Which would get you cos(θ) / cos(θ)2 * tan(θ) And then coming out to 1 / cos(θ) *...

could someone explain finding cos(11pi/12) using Sum and difference Identities. it must use one of these: cos(α+β)=cos(α)cos(β)−sin(α)sin(β) or cos(α+β)=cos(α)cos(β)+sin(α)sin(β) Im...

Solve the given equation over the interval [0,2): 2sinx+ (sqrt 3) sin^2x=-(sqrt 3) cos^2x A. x=0, and x=(pi/2), and x=(3 pi/2), and x=2(pi)x=(3 pi/2), and x=(7 pi/2) b. x=(4 pi/3) and...

cos7y cos3y -sin 7y sin3y= ?

i need step by step soln plz thank u

Please help me solve this equation to better help me to understand trig Identites.

I am a little confused on this and it would be nice if someone could work through it so I can understand this problem

1) (cscx + cotx) / (tanx + sinx) = cotxcscx 2) (1/secx) -secx = -sinx 3) (1-tan^2x / 1-cot^2x) = -tan^2x

=(sinx/cosx - sinx)/ (2sinx/cosx) = sinx-sinxcosx/cosx • cosx/2sinx = sinx (1-cosx)/2sinx = 1-cosx/2 now im lost help!!

Using only the double angle identities how to solve for this? im stucked after 2cos^2 (2x) -1

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.

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

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

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...

(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