If -4-2- then write the value of -1-sin 2 -

Since π/4 lt; θ lt; π/2 ⇒ θ lies in the first quadrant 1^st quadrant Now, √(1 sin 2 θ) =√(sin^2 θ + cos^2 θ 2 sin θ cos θ) = √((sin θ cos θ)^2) = ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

If π/4<θ<π/2,...

Question

If $\frac{π}{4} < θ < \frac{π}{2}$ , then write the value of $\sqrt{1 - s i n 2 θ}$

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Solution

Since $\frac{π}{4} < θ < \frac{π}{2}$

$\Rightarrow θ$ lies in the first quadrant $1^{s t}$ quadrant

Now,

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

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If π4<θ<π2, then write the value of √1−sin 2θ

Since π4<θ<π2 ⇒ θ lies in the first quadrant 1st quadrant Now, √1−sin 2θ=√sin2 θ+cos2 θ−2 sin θ cos θ=√(sin θ−cos θ)2=sin θ−cos θ

If $\frac{π}{4} < θ < \frac{π}{2}$ , then write the value of $\sqrt{1 - s i n 2 θ}$

Since $\frac{π}{4} < θ < \frac{π}{2}$

$\Rightarrow θ$ lies in the first quadrant $1^{s t}$ quadrant

Now,

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