1) (cscx + cotx) / (tanx + sinx) = cotxcscx 2) (1/secx) -secx = -sinx 3) (1-tan^2x / 1-cot^2x) = -tan^2x
1) (cscx + cotx) / (tanx + sinx) = cotxcscx 2) (1/secx) -secx = -sinx 3) (1-tan^2x / 1-cot^2x) = -tan^2x
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...
=(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...
tan34/2(1-tan34) =(2/2)[tan34/2(1-tan34) =2tan34/4(1-tan34) i need to have the form 2tanA/1-tan^2A but i cant eliminate the 4
tana+tanb/1-tanatanb im confused becase sqrt of 3 over 3 is the value of tan 30 so its hard to solve the arithmetic and the algebra. help please!
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.
Please help, I have no idea! Solve the following trigonometric equation sin^2x - cos^2x = 1/4.
(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
tan3x - tan2x- tanx = tan3xtan2xtanx
sin(X)sec(x)=1-(cos^2)x/sin(X)cos(X)
This is for my trig class. Thanks for the help!
This is for my trig class. Thanks for the help!
This is for my trig class, and the directions say to verify the identity. Thanks so much.
1) given the double angle identity for cosine: cos2θ=cos2θ-sin2θ Prove: cos3θ = cos3θ - 3cosθsin2θ and: cos4θ = cos4θ - 6cos2θsin2θ + sin2θ 2)...
(1-cos^2x)(cot^2+1) = 1