If sin Θ = negative square root 3 over 2 and π < Θ < 3 pi over 2, what are the values of cos Θ and tan Θ?

Answer:  The values are

[tex]\cos\theta=-\dfrac{1}{2},~~\textup{and}~~\tan\theta=\sqrt3.[/tex]

Step-by-step explanation:  For an angle [tex]\theta[/tex],

[tex]\sin \theta=-\dfrac{\sqrt3}{2},~\pi<\theta<\dfrac{3\pi}{2}.[/tex]

We are given to find the values of [tex]\cos\theta[/tex] and [tex]\tan \theta[/tex].

Given that

[tex]\pi<\theta<\dfrac{3\pi}{2}\\\\\\\Rightarrow \theta~\textup{lies in Quadrant III}.[/tex]

We will be using the following trigonometric properties:

[tex](i)~\cos \theta=\pm\sqrt{1-\sin^2\theta},\\\\(ii)~\tan\theta=\dfrac{\sin\theta}{\cos\theta}.[/tex]

The calculations are as follows:

We have

[tex]\cos\theta=\pm\sqrt{1-\sin^2\theta}\\\\\\\Rightarrow \cos \theta=\pm\sqrt{1-\left(-\dfrac{\sqrt3}{2}\right)^2}\\\\\\\Rightarrow \cos\theta=\pm\sqrt{1-\left(\dfrac{3}{4}\right)}\\\\\\\Rightarrow \cos\theta=\pm\sqrt{\left(\dfrac{1}{4}\right)}\\\\\\\Rightarrow \cos\theta=\pm\dfrac{1}{2}.[/tex]

Since [tex]\theta[/tex] is in Quadrant III, and the value of cosine is negative in that quadrant, so

[tex]\cos\theta=-\dfrac{1}{2}.[/tex]

Now, we have

[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{\sqrt3}{2}}{-\frac{1}{2}}=\sqrt3.[/tex]

Thus, the values are

[tex]\cos\theta=-\dfrac{1}{2},~~\textup{and}~~\tan\theta=\sqrt3.[/tex]

Since the given theta lies in third quadrant, then you can use the fact that only tangent and cotangent are positive in third quadrant, rest are negative.

The value of cos and tan for given theta is:

[tex]cos(\theta) = -\dfrac{1}{2}\\\\ tan( \theta) = \sqrt{3}[/tex]

If angle lies between 0 to half of pi, then it is int first quadrant.

If angle lies between half of pi to a pi, then it is in second quadrant.

When the angle lies between [tex]\pi[/tex] and [tex]\dfrac{3\pi}{2}[/tex], then that angle lies in 3rd quadrant.

And when it lies from [tex]\dfrac{3\pi}{2}[/tex] and 0 degrees, then the angle is in fourth quadrant.

Only tangent function and cotangent functions.

In first quadrant, all six trigonometric functions are positive.

In second quadrant, only sin and cosec are positive.
In fourth, only cos and sec are positive.

We can continue as follows:

[tex]sin(\theta) = -\dfrac{\sqrt{3}}{2}\\ sin(\theta) = sin(\pi + 60^\circ)\\ \theta = \pi + 60^\circ[/tex]

Thus, evaluating cos and tan at obtained theta:

[tex]cos(\pi + 60^\circ) = -cos(60) = -\dfrac{1}{2}\\ tan( \pi + 60^\circ) = tan(60) = \sqrt{3}[/tex]

Learn more about trigonometric functions and quadrants here;

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