Question

Angle θ is in standard position. if sin(θ) = − 1/3, and π < θ < 3π/2 , find cos(θ).
A. -(2√2)/3
B.4/3
C.2√2/3

Answer 1

Use pythagorean theorem:
sin^2 + cos^2 = 1

(-1/3)^2 + cos^2 = 1
1/9 + cos^2 = 1
cos^2 = 8/9
cos = +- sqrt(8)/sqrt(9) = +- 2sqrt(2)/3

Determine whether cos is positive or negative by looking at which quadrant the angle is in.
pi < theta < 3pi/2 ---> this is 3rd quadrant where x or cos is negative

Therefore cos(theta) = -2sqrt(2)/3

Answer 2

You can use the first Pythagorean identity along with the fact that cosine is not positive in third quadrant.

The value of cos(θ) in given context is given by

A. -(2√2)/3

What are the three Pythagorean identities ?

$sin^2(\theta) + cos^2(\theta) = 1\\\\1 + tan^2(\theta) = sec^2(\theta)\\\\1 + cot^2(\theta) = csc^2(\theta)$

Which trigonometric functions are positive in which quadrant?

• In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
• In second quadrant(π/2 < θ < π), only sin and cosec are positive.
• In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
• In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle  as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Using above facts to find the value of cos(θ):

Since the given angle is in third quadrant, we have cos(θ) negative.

Now, using the first Pythagorean identity, we get:

$sin^2(\theta) + cos^2(\theta) = 1\\\\cos(\theta) = \pm \sqrt{1 - sin^2(\theta)}$

we will take negative sign as concluded above. Thus,

$cos(\theta) = - \sqrt{1 - sin^2(\theta)}$

Now, since sin(θ) = − 1/3, thus,

$cos(\theta) = - \sqrt{1 - sin^2(\theta)} = -\sqrt{1 - (-1/3)^2} = -\sqrt{\dfrac{9-1}{9}} = -\dfrac{2\sqrt{2}}{3}$

Thus,

The value of cos(θ) in given context is given by

A. -(2√2)/3

Learn more about first Pythagorean identity here:

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