Find the exact value. sin 𝜽 = -2⎷13/13 where 3𝛑/2 ≤ 𝜽 < 2𝛑 Find tan2𝜽

Question

I am confused and would appreciate some help with work showing please.

Answer 1

sin 𝜽 = -2⎷13/13

" " = -2*3.605/13=.5547...........

θ=-33.69° or 146°+180=326°

3𝛑/2 ≤ 𝜽 < 2𝛑 Find tan2𝜽

540/2

270 " 360

Answer 2

From the description, the angle θ is in the forth quadrant. So tangent will be negative and cosine positive.

Use the double angle formula for sine and cosine.

tan(2θ) = 2sin(θ)cos(θ)/(cos²(θ) - sin²(θ))

We know sin(θ) = -2√13/13 = opposite/hypotenuse = y/h

We need cos(θ) = adjacent/hypotenuse = x/h

x = √(13² - (-2√13)²) = √(169 - 52) = 3√13

cos(θ) = 3√13/13

tan(2θ) = 2(-2√13/13)(3√13/13) /((3√13/13)² - (-2√13/13)²)

= (-12(13)/169)/(5(13)/169) = -12/5

Or

We could use

tan(2θ) = 2tan(θ)/(1-tan²(θ))

tan(θ) = -(2√13)/(3√13) = -2/3

tan(2θ) = -(4/3)/(1-4/9) = -(4/3)(9)/(9-4) = -12/5

Which is much easier.