Asked by Zoë
Solve the following trigonometry identities.
a) 1-cos2(theta) = sin(theta)cos(theta)/cot(theta)
b) (1-cos2(theta))(1-tan2(theta))=sin2(theta)-2sin4(theta)/1-sin2(theta)
*its supposed to be cos to the power of two, sin to the power of four, etc. There is also supposed to be a theta symbol where the word is.
a) 1-cos2(theta) = sin(theta)cos(theta)/cot(theta)
b) (1-cos2(theta))(1-tan2(theta))=sin2(theta)-2sin4(theta)/1-sin2(theta)
*its supposed to be cos to the power of two, sin to the power of four, etc. There is also supposed to be a theta symbol where the word is.
Answers
Answered by
Steve
you don't solve an identity -- you prove it.
1-cos^2θ = sinθ cosθ/cotθ
since cotθ = cosθ/sinθ, now we have
sin^2θ = sinθ sinθ
QED
(1-cos^2θ)(1-tan^2θ) = (sin^2θ-2sin^4θ/(1-sin^2θ)
sin^2θ(1-tan^2θ) = sin^2θ(1-2sin^2θ)/(1-sin^2θ)
sin^2θ(1-tan^2θ) = sin^2θ(1-2sin^2θ)/cos^2θ
sin^2θ(1-tan^2θ) = tan^2θ(1-sin^2θ-sin^2θ)
sin^2θ(1-tan^2θ) = tan^2θ(cos^2θ-sin^2θ)
sin^2θ(1-tan^2θ) = sin^2θ(1-tan^2θ)
1-cos^2θ = sinθ cosθ/cotθ
since cotθ = cosθ/sinθ, now we have
sin^2θ = sinθ sinθ
QED
(1-cos^2θ)(1-tan^2θ) = (sin^2θ-2sin^4θ/(1-sin^2θ)
sin^2θ(1-tan^2θ) = sin^2θ(1-2sin^2θ)/(1-sin^2θ)
sin^2θ(1-tan^2θ) = sin^2θ(1-2sin^2θ)/cos^2θ
sin^2θ(1-tan^2θ) = tan^2θ(1-sin^2θ-sin^2θ)
sin^2θ(1-tan^2θ) = tan^2θ(cos^2θ-sin^2θ)
sin^2θ(1-tan^2θ) = sin^2θ(1-tan^2θ)
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