Question

At what point does the terminal side of the angle 5π6 in standard position intersect the unit circle?

(-3√2, 12)

(-12, 3√2)

(12, −3√2)

(3√2, −12)


Which point on the unit circle corresponds to −π6 ?

(-12, 3√2)

(3√2, −12)

(12, −3√2)

(-3√2, 12)


What is the exact value of sin(5π4) ?

−12

2√2

12

−2√2


What is the exact value of tan(−3π4) ?

Enter your answer in the box.

______

In which quadrant does θ lie given that sinθ<0 and cosθ>0 ?

Quadrant I

Quadrant II

Quadrant III

Quadrant IV

Answer 1

Answer:

1.) At what point does the terminal side of the angle 5π/6 in standard position intersect the unit circle?

First option: ( -√3/2, 1/2)

2. Which point on the unit circle corresponds to −π/6 ?

Second option: ( √3/2, -1/2)

3. What is the exact value of sin(5π/4) ?

Fourth option: −√2/2

4. What is the exact value of tan(−3π/4) ?

tan(−3π/4) = 1

5. In which quadrant does θ lie given that sinθ<0 and cosθ>0 ?

Fourth option: Quadrant IV

Step-by-step explanation:

1.) At what point does the terminal side of the angle 5π6 in standard position intersect the unit circle?

Point: P=(x,y)=?

x=cos 5π/6 → x = -√3/2

y=sin 5π/6 → y = 1/2

P=( -√3/2, 1/2)

2. Which point on the unit circle corresponds to −π/6 ?

Point: P=(x,y)=?

x=cos (-π/6) → x = √3/2

y=sin (-π/6) → y = -1/2

P=( √3/2, -1/2)

3. What is the exact value of sin(5π/4) ?

sin(5π/4) = −√2/2

4. What is the exact value of tan(−3π/4) ?

tan(−3π/4) = 1

5. In which quadrant does θ lie given that sinθ<0 and cosθ>0 ?

sinθ<0 (-) in Quadrants III and IV

cosθ>0 (+) in Quadrants I and IV

Then, sinθ<0 and cosθ>0 in Quadrant IV

Answer 2

1) $\rm P = (-\dfrac{\sqrt{3} }{2},\dfrac{1}{2})$

2) $\rm Q =(\dfrac{\sqrt{3} }{2},-\dfrac{1}{2})$

3) $sin(\dfrac{5\pi}{6}) = -\dfrac{1}{\sqrt{2} }$

4) $tan(-\dfrac{3\pi}{4})=1$

5) The correct answer is Quadrant IV.

Step-by-step explanation:

1) Let the point be P(x,y)

$x = cos (\dfrac{5\pi}{6})=-\dfrac{\sqrt{3} }{2}$

$y = sin (\dfrac{5\pi}{6})=\dfrac{1}{2}$

$\rm P = (-\dfrac{\sqrt{3} }{2},\dfrac{1}{2})$

2) Let that point be Q(x,y)

$x=cos(-\dfrac{\pi}{6})=\dfrac{\sqrt{3} }{2}$

$y=sin(-\dfrac{\pi}{6})=-\dfrac{1}{2}$

$\rm Q =(\dfrac{\sqrt{3} }{2},-\dfrac{1}{2})$

3)

$sin(\dfrac{5\pi}{6}) = -\dfrac{1}{\sqrt{2} }$

4)

$tan(-\dfrac{3\pi}{4})=1$

5) $sin\theta < 0$ in quadrants III and IV

    $cos\theta > 0$ in quadrants I and IV

Therefore the correct answer is Quadrant IV.

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https://brainly.com/question/20226319?referrer=searchResults