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Jacobbc2
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Jul 15 2021 12:11am
My brain is literally shutting down rn and i'm on the last problem of my homework

If anyone remembers trig it'd help a ton

I'm trying to solve for secant(a) if sin(a)=-2/3 and a is in the 4th quadrant

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Jul 16 2021 12:02am
1) draw a right triangle in the 4th quadrant and label the acute angle formed with the horizontal axis a. Please note that in the 4th quadrant your horizontal component of the triangle is positive and your vertical component is negative. Hypotenuse is always positive (any quadrant).

2) use the given trig function and label the sides of the triangle.

Sin(a) = -2 / 3

Sin(a) = opposite / hypotenuse

You can label the opposite side -2 and the hypotenuse as 3

3) you’ll likely have to solve for the missing side using the Pythagorean theorem.

The missing side is the adjacent side and is + sqrt(5). This side must be positive since your triangle is in quadrant 4 where the horizontal value is positive.

4) the triangle is labeled at this point and you can actually find ALL the trigonometric functions now. We only care about sec(a) though.

Sec(a) is really 1/cos(a). <—- find cos(a) then flip the fraction.

Cos(a) = sqrt(5) / 3

This means

Sec(a) = 3 / sqrt(5)

5) depending on your instructor they will likely have you rationalize any denominators. This means no sqrt in the bottom of the fraction.

We can multiply the fraction by sqrt(5)/sqrt(5) to achieve this.

Sec(a) = ( 3 sqrt(5) ) / ( sqrt(5) sqrt(5) )

Simplify the denominator to get

Sec(a) = ( 3 sqrt(5) ) / 5
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