Question

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Answer 1

Given

sin α=-3/5 where π< α< 3π/2

To find

The value of cos(α/2)

Solution :

We have to find the value of $\sf\cos(\dfrac{\alpha}{2})$

Given :

$\sf\sin\alpha=\dfrac{-3}{5}$

We know that,sin² x+cos²x=1 then

$\sf\cos\alpha=\sqrt{1-\sin^2\alpha}$

$\sf\cos\alpha=\sqrt{1-(\dfrac{-3}{5}})^2$

$\sf\cos\alpha=\sqrt{\dfrac{16}{25}}$

$\sf\cos\alpha=\pm\dfrac{4}{5}$

Since,π< α< 3π/2

It means ,cosα lies in third quadrant ,in third quadrant cos is negative .

Then,$\sf\cos\alpha=\dfrac{-4}{5}..(1)$

We know that

$\sf\cos2\alpha=2\cos^2\alpha-1$

$\sf\cos\alpha=2\cos^2(\dfrac{\alpha}{2})-1$

$\sf\cos(\dfrac{\alpha}{2})=\sqrt{\dfrac{\cos\alpha+1}{2}}$

From equation (1)

$\sf\cos(\dfrac{\alpha}{2})=\sqrt{\dfrac{\frac{-4}{5}+1}{2}}$

$\sf\cos(\dfrac{\alpha}{2})=\sqrt{\dfrac{-4+5}{10}}$

$\sf\cos(\dfrac{\alpha}{2})=\dfrac{1}{\sqrt{10}}$

Hence , Correct option is b) cosα=1/√10

Answer 2

Answer:

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