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frutty [35]
3 years ago
13

Given: cos θ=-4/5, sin x = -12/13, θ is in the third quadrant, 

Mathematics
1 answer:
USPshnik [31]3 years ago
4 0

By definition of tangent,

tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)

Recall the double angle identities:

sin(2<em>θ</em>) = 2 sin(<em>θ</em>) cos(<em>θ</em>)

cos(2<em>θ</em>) = cos²(<em>θ</em>) - sin²(<em>θ</em>) = 2 cos²(<em>θ</em>) - 1

where the latter equality follows from the Pythagorean identity, cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1. From this identity we can solve for the unknown value of sin(<em>θ</em>):

sin(<em>θ</em>) = ± √(1 - cos²(<em>θ</em>))

and the sign of sin(<em>θ</em>) is determined by the quadrant in which the angle terminates.

<em />

We're given that <em>θ</em> belongs to the third quadrant, for which both sin(<em>θ</em>) and cos(<em>θ</em>) are negative. So if cos(<em>θ</em>) = -4/5, we get

sin(<em>θ</em>) = - √(1 - (-4/5)²) = -3/5

Then

tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)

tan(2<em>θ</em>) = (2 sin(<em>θ</em>) cos(<em>θ</em>)) / (2 cos²(<em>θ</em>) - 1)

tan(2<em>θ</em>) = (2 (-3/5) (-4/5)) / (2 (-4/5)² - 1)

tan(2<em>θ</em>) = 24/7

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<em><u>Solution:</u></em>

Given that, a rectangle with a length of 5 and a width of 3x - 5y + 6

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