If -2- then write the value of -2-2-2 cos 2 - in the simplest form-

Since π/2 lt; θ lt; π ⇒ θ lies in the 2^nd quadrant Now, √(2 ±√(2+2 cos 2 θ)) =√(2 ±√(2(1+cos 2 θ))) = √(2 ±√(2.2 cos^2 θ)) =√(2 (2 cos θ)) (∵ θ lies in ...

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Standard XII

Mathematics

Trigonometric Ratios of Multiple of an Angle

If π/2<θ<π, t...

Question

If $\frac{π}{2} < θ < π$ , then write the value of $\sqrt{2 + \sqrt{2 + 2 c o s 2 θ}}$ in the simplest form.

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Solution

Since $\frac{π}{2} < θ < π$

$\Rightarrow$ $θ$ lies in the $2^{n d}$ quadrant

Now,

$\sqrt{2 \pm \sqrt{2 + 2 c o s 2 θ}} = \sqrt{2 \pm \sqrt{2 (1 + c o s 2 θ)}} = \sqrt{2 \pm \sqrt{2.2 c o s^{2} θ}} = \sqrt{2 - (2 c o s θ)}$

( $∵ θ$ lies in the $2^{n d}$ quadrant)

$= \sqrt{2 (1 - c o s θ)} = 2.2 s i n^{2} \frac{θ}{2} = 2 s i n \frac{θ}{2}$

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If π2<θ<π, then write the value of √2+√2+2 cos 2 θ in the simplest form.

Since π2<θ<π ⇒ θ lies in the 2nd quadrant Now, √2±√2+2 cos 2 θ=√2±√2(1+cos 2θ)=√2±√2.2 cos2 θ=√2−(2 cos θ) (∵ θ lies in the 2nd quadrant) =√2(1−cos θ)=2.2 sin2 θ2=2 sin θ2

If $\frac{π}{2} < θ < π$ , then write the value of $\sqrt{2 + \sqrt{2 + 2 c o s 2 θ}}$ in the simplest form.