(secθ-tanθ)^2=2secθtanθ
(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)...
2sin^2x - 1÷ sinxcosx = tanx-cotx
(1-cos^2x)(cot^2+1) = 1
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simplify simplify simplify simplify ...............
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