Asked by Bri
How would you write cos(tan^-1u) as an algebraic expression in u?
I know the answer, and I've seen tutorial on how to do it using pathagorus, but how would you do this using identities?
I know the answer, and I've seen tutorial on how to do it using pathagorus, but how would you do this using identities?
Answers
Answered by
Steve
draw the triangle
In the triangle, label the sides 1 and u, and the hypotenuse √(1+u^2)
the angle tan^-1(u) is opposite the "u" side, since opposite/adjacent = u/1 = u
So, the cosine of that angle is adjacent/hypotenuse = 1/√(1+u^2)
You can also note that
cosθ = 1/secθ
sec^2θ = 1+tan^2θ, so
cosθ = 1/√(1+tan^2 θ)
But θ = tan^-1(u), so tanθ = u and we have
cosθ = 1/√(1+u^2)
I have always easier to carefully draw the triangle and label the sides. Then all the trig functions just drop right out.
In the triangle, label the sides 1 and u, and the hypotenuse √(1+u^2)
the angle tan^-1(u) is opposite the "u" side, since opposite/adjacent = u/1 = u
So, the cosine of that angle is adjacent/hypotenuse = 1/√(1+u^2)
You can also note that
cosθ = 1/secθ
sec^2θ = 1+tan^2θ, so
cosθ = 1/√(1+tan^2 θ)
But θ = tan^-1(u), so tanθ = u and we have
cosθ = 1/√(1+u^2)
I have always easier to carefully draw the triangle and label the sides. Then all the trig functions just drop right out.
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