Trigonometry Practice Test 1

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Question : 7 of 40

Marks: +1, -0

If sin θ =

- 3/5

and θ lies in the third quadrant, then the value of

\cos (θ/2)

is:

$1/5$
$-1/√(10)$
$-1/5$
$1/√(10)$

Solution:

∵ θ lies in the third quadrant
⇒ cos θ is –ve
⇒ sin θ =

- 3/5

⇒ cosθ =

- √(1- \sin^2 θ) = - 4/5

Now,

\cos^2 1/2θ=1/2(1+\cos θ)

=1/2(1-4/5)=1/10

⇒\cos 1/2θ=±1/√10

But given that, 180° < θ < 270°

⇒ 90° < θ/2 < 135°

i.e.,

θ/2

lies in the second quadrant

⇒\cos 1/2θ

is -ve
Hence

\cos 1/2θ=-1/√10

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