If sinθ = -4/5 and θ Lies in the Third Quadrant, Find cos(θ/2)

If \( \sin\theta=-\frac45 \) and \( \theta \) Lies in the Third Quadrant, Find \( \cos\frac{\theta}{2} \)

Question

If

\[ \sin\theta=-\frac45 \]

and \(\theta\) lies in the third quadrant, then the value of

\[ \cos\frac{\theta}{2} \]

is

(a) \(\frac15\)
(b) \(\frac1{\sqrt{10}}\)
(c) \(\frac1{\sqrt5}\)
(d) \(\frac{3}{\sqrt{10}}\)

Solution

Given

\[ \sin\theta=-\frac45 \]

Since \(\theta\) is in the third quadrant, \(\cos\theta\) is also negative.

Using

\[ \sin^2\theta+\cos^2\theta=1 \]

\[ \cos\theta = -\sqrt{1-\left(\frac45\right)^2} = -\sqrt{\frac{9}{25}} = -\frac35 \]

Now use the half-angle identity:

\[ \cos\frac{\theta}{2} = \pm\sqrt{\frac{1+\cos\theta}{2}} \]

Since \(\theta\) is in the third quadrant,

\[ 180^\circ<\theta<270^\circ \]

Therefore,

\[ 90^\circ<\frac{\theta}{2}<135^\circ \]

So \(\frac{\theta}{2}\) lies in the second quadrant, where cosine is negative.

Hence,

\[ \cos\frac{\theta}{2} = -\sqrt{\frac{1-\frac35}{2}} \]

\[ = -\sqrt{\frac{\frac25}{2}} = -\sqrt{\frac15} \]

\[ = -\frac1{\sqrt5} \]

Final Answer

\[ \boxed{-\frac1{\sqrt5}} \]

The exact value is \(\boxed{-\frac1{\sqrt5}}\). Since this is not listed among the given options, the question appears to have a sign error in the options. The magnitude corresponds to (c) \(\frac1{\sqrt5}\).

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