If cosx=−5/13 and tanx<0, then find sin2x

Question

I need to find the exact functional value

Answer 1

If the cosx = negative and tanx = negative

Then, the angle x is in Q2

In Q2, the cos is negative, sin is positive

Construct a right triangle with cosX = 5/13

Using Pythagorean Theorem:

x^2 + y^2 = c^2

5^2 + y^2 = 13^2

y = sqrt [ 13^2 - 5^2 ]

y = sqrt [ 169 - 25 ]

y = sqrt [ 144 ] = 12

Therefore, sinX = 12/13

Use Double Angle Formula

sin(2x) = 2 * sinX * cosX

sin(2x) = 2 * (12/13) * (-5/13)

sin(2x) = (-120 /169) Exact Answer

sin(2x) = - .71 Approx. Answer

CHECK:

x = ArcCos (-5/13) = 112.62 deg.

sin(2x) = sin(225.24) = -.71

Answer 2

sin2x = 2sinxcosx

sinx^2 + cosx^2 = 1

sinx^2 + (-5/13)^2 = 1

sinx^2 = 144/169

sinx = +/- 12/13

if tanx < 0 and cosx is negative, sinx must be positive and equal to 12/13

so sin2x = 2 * 12/13 * -5/13 = -120/169 = -0.71