If tan x = (2ab)/(a^2 - b^2), where a>b>0 and x is between zero and ninety degrees, then find sin x in terms of a and b.
I changed the tan x to (sin x)/(cos x), but I have no idea what to do next. Please explain all of the steps. Thank you so much!
I changed the tan x to (sin x)/(cos x), but I have no idea what to do next. Please explain all of the steps. Thank you so much!
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Since 0° < x < 90°, x is in Quadrant I and sin(x) > 0.
By drawing a triangle in Quadrant I such that tan(x) = opp/adj = 2ab/(a^2 - b^2), we have:
opp = 2ab, adj = a^2 - b^2, and hyp = √[(2ab)^2 + (a^2 - b^2)^2] = a^2 + b^2.
(Note that (2ab)^2 + (a^2 - b^2)^2 = a^4 + 2a^2*b^2 + b^4 = (a^2 + b^2)^2.)
Therefore:
sin(x) = opp/hyp = 2ab/(a^2 + b^2).
I hope this helps!
By drawing a triangle in Quadrant I such that tan(x) = opp/adj = 2ab/(a^2 - b^2), we have:
opp = 2ab, adj = a^2 - b^2, and hyp = √[(2ab)^2 + (a^2 - b^2)^2] = a^2 + b^2.
(Note that (2ab)^2 + (a^2 - b^2)^2 = a^4 + 2a^2*b^2 + b^4 = (a^2 + b^2)^2.)
Therefore:
sin(x) = opp/hyp = 2ab/(a^2 + b^2).
I hope this helps!