Sam Z. answered • 02/24/21

Math/Science Tutor

**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

Sofia D.

asked • 02/24/21I am confused and would appreciate some help with work showing please.

Follow
•
1

Add comment

More

Report

Sam Z. answered • 02/24/21

Tutor

4.3
(12)
Math/Science Tutor

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

θ=-33.69° or 146°+180=326°

540/2

270 " 360

Bradford T. answered • 02/24/21

Tutor

4.9
(8)
MS in Electrical Engineering with 40+ years as an Engineer

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)/(cos^{2}(θ) - sin^{2}(θ))

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

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

x = √(13^{2} - (-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-tan^{2}(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.

Ask a question for free

Get a free answer to a quick problem.

Most questions answered within 4 hours.

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.