Question

If cosine theta equals negative square root of three over two and pi over two less than theta less than pi, what are the values of sin Θ and tan Θ?
sine theta equals negative one half, tangent theta equals square root of three over three
sine theta equals negative one half, tangent theta equals negative square root of three over three
sine theta equals one half, tangent theta equals negative square root of three over three
sine theta equals one half, tangent theta equals square root of three over three

Answer 1

Answer:

The values of sinθ and tanθ are:

$\sin \theta=\dfrac{1}{2}$

and

$\tan \theta=-\dfrac{1}{\sqrt{3}}=-\dfrac{\sqrt{3}}{3}$

Step-by-step explanation:

It is given that:

$\cos \theta=-\dfrac{\sqrt{3}}{2}$

where

$\dfrac{\pi}{2}$

i.e. the angle θ lie in the second quadrant.

Also, we know that in the second quadrant sine and cosecant function is positive.

whereas cosine,secant, tangent and cotangent functions are negative.

Now, we know that:

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

i.e. here

$\theta=\dfrac{5\pi}{6}$

Hence, we have:

$\sin (\theta)=\sin (\dfrac{5\pi}{6})\\\\i.e.\\\\\sin \theta=\dfrac{1}{2}$

Also,

$\tan \theta=\dfrac{\sin \theta}{\cos \theta}\\\\i.e.\\\\\tan \theta=\dfrac{\dfrac{1}{2}}{\dfrac{-\sqrt{3}}{2}}\\\\i.e.\\\\\tan \theta=\dfrac{1}{-\sqrt{3}}\\\\i.e.\\\\\tan \theta=-\dfrac{1}{\sqrt{3}}$

Now, on rationalizing we have:

$\tan \theta=-\dfrac{\sqrt{3}}{3}$

Answer 2

Answer:

sin θ = 1/2

tan θ = -√3 / 3

Step-by-step explanation:

cos θ = -√3 / 2, π/2 < θ < π

θ is in the second quadrant, so θ = 5π/6.

sin θ = 1/2

tan θ = sin θ / cos θ = -√3 / 3