Question

Given sec⁡θ=-11/3,180°<θ<270°, find sin θ

Answer 1

Answer:
`-(4√(7))/(11)`

==================================================

Work Shown

180° < θ < 270° represents quadrant Q3 (southwest quadrant).

sec⁡θ = -11/3

cos⁡θ = -3/11 since secant is the reciprocal of cosine, and vice versa.

Use the pythagorean trig identity to find sin⁡θ

`\sin^2(\theta)+\cos^2(\theta) = 1\\\\\sin(\theta) = -√(1-\cos^2(\theta)) \ \text{ ... sine is negative in Q3}\\\\\sin(\theta) = -\sqrt{1-\left(-(3)/(11)\right)^2}\\\\\sin(\theta) = -\sqrt{(112)/(121)\\\\`

`\sin(\theta) = -(√(112))/(√(121))\\\\\sin(\theta) = -(√(16*7))/(√(11^2))\\\\\sin(\theta) = -(√(16)*√(7))/(11)\\\\\sin(\theta) = -(4√(7))/(11)\\\\`

The CAS (computer algebra system) feature of GeoGebra can be used to verify the answer. WolframAlpha is another option.

----------------------------

Another approach:

secant is the ratio of hypotenuse over adjacent. It's the reciprocal of cosine = adjacent/hypotenuse.

Since sec⁡θ = -11/3, we know that hypotenuse = 11 and adjacent = -3.

Use the pythagorean theorem to find that opposite =
`-4√(7)`

Then use the ratio sine = opposite/hypotenuse to get the answer shown above.

I made adjacent and opposite negative to help remember the proper signs.