If tan x= 12/5, and 0º<x<90º, whats sin(2x)?

Question

Hi,

tan(x) = 12/5

as we know: 1 + tan(x) ² = sec²(x)

1 + (12/5) (12/5) = sec²x

1 + 144/25 = sec²x

25/25 + 144/25 = sec²x

169/25 = sec²x

now take a root of both side

13/5 =± sec x must choose positive because the angle, x, is first quadrant

as you know sec x and cos x are reciprocal of each other sec x =1 / cos x

cos x = 5/13

on the other hand. cos x= adj / hyp as a result adj = 5 and hyp = 13

we know 13² = adj² + opp²

169 = 5² + opp²

169 = 25 + opp2

^{169 - 25 = opp2}

144 = opp² as a result opp = ±12 because of first quadrant opp = 12

now you know sin x = 12/13

so sin x = 12/13 and cos x = 5/13

as you know sin 2x = 2 sin x cos x

sin 2x = 2 * 12/13 * 5/13

sin 2x = 120 /169

I hope it is helpful,

Feel free if you have any question

so cos x = 5 / 13 ±

Answer 1

That works, but there is a quicker and understandable route to the answer:

tan x = opp/adj = 12/5

hyp^2 = opp^2 + adj^2 = 12^2 + 5^2 = 144 + 15 = 169 = 13^2 ==> hyp =13

sin x = opp/hyp = 12/13

cos x = adj/hyp = 5/13

cos x, sin x, and tan x are all positive in 0 <= x <= 90 degrees (first quadrant).

sin 2x = 2 sin x cos x = 2 (12/13)(5/13) = 120/169 as expressed

[Technically, opp = 12C, adj = 5C, hyp^2 = (13C)^2, and hyp = 13C where C is a constant.]