Asked by dam
express root(9+x^2) as a trigonomentric function of tita without using radicals by letting x=3tantita.assume that 0<tita<TT/2 hence find sine tita and cos tita
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Answers
Answered by
Steve
theta, not tita.
let x = 3tanθ
x^2 = 9tan^2θ
9+x^2 = 9+9tan^2θ = 9sec^2θ
√(9+x^2) = 3secθ
so, draw a right triangle. Since tanθ = x/3, those are the two legs, and the hypotenuse is √(9+x^2)
Now you can write down sinθ and cosθ
let x = 3tanθ
x^2 = 9tan^2θ
9+x^2 = 9+9tan^2θ = 9sec^2θ
√(9+x^2) = 3secθ
so, draw a right triangle. Since tanθ = x/3, those are the two legs, and the hypotenuse is √(9+x^2)
Now you can write down sinθ and cosθ
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