Question

11th Warm Up ‼️ Answer this Question with Solution ​

Answer 1

Solution :

$\sf\sqrt{2+\sqrt{2+2\cos4\theta}}=2\sin\theta$

Step by step Explanation :

Given : π < 2θ < 3π/2

We have to find the value of $\sf\sqrt{2+\sqrt{2+2\cos4\theta}}$

We know that

cos2θ=2cos²θ-1

Then , cos4θ=2(cos2θ)²-1

➝ 1+cos4θ=2(cos2θ)²

Now ,

$\sf\sqrt{2+\sqrt{2+2\cos4\theta}}$

$\sf=\sqrt{2+\sqrt{2(1+\cos4\theta)}}$

$\sf=\sqrt{2+\sqrt{2(2\cos^{2}2\theta)}}$

$\sf=\sqrt{2+\sqrt{4cos^{2}2\theta}}$

Here,

$\sqrt{4cos^{2}2\theta}=\pm2\cos2\theta$

Now we will see that what θ mean here to us . As given in the equation

π < 2θ < 3π/2

It signifies that 2θ lies in the third Quadrant i.e 180° < 2θ < 270°

We know that : In third Quadrant cosine is -ve .

Now,

$\sf\sqrt{2+\sqrt{4cos^{2}2\theta}}$

$\sf=\sqrt{2-2cos2\theta}$

$\sf=\sqrt{2(1-cos2\theta)}$

$\sf=\sqrt{2(2\sin^{2}\theta)}$

$\sf=\sqrt{4\sin^2\theta}$

$\sf=\pm2\sin\theta$

Now we'll find about θ

180° < 2θ < 270°

➝ 90° < θ < 135°

Thus , θ lies in second quadrant , where sin function is +ve .

Therefore, Correct option d)2sinθ

Some More Formula's :

1. sin2A = 2 sinA cosA
2. cos2A = cos²A - sin²A
3. tan2A = 2 tanA / (1 - tan²A)
4. cos2θ=2cos²θ-1
5. cos2θ=1-2sin²θ
6. sin2θ=2sinθcosθ

Answer 2

Solution:-

$\sqrt{ 2 + \sqrt{2 + 2 \cos(4 \theta) } } \: = \: 2 \sin( \theta)$

Step by Step explanation:

Given : π < 2$\sf{theta}$<3π/2

We know that

Cos 2$\sf{theta}$=

$\sf \: 2 { ( cos) }^{2} \theta \: - 1 \\ \\ \sf \: then \: \cos(4 \theta) = 2 {(cos2 \theta)}^{2} - 1 \\ \\ \implies \: 1 + \cos(4 \theta) = 2 {(cos2 \theta)}^{2}$

Now,

$\sf \sqrt{2 + \sqrt{2 + 2 \cos(4 \theta) } } \\ \\ \sf \implies \: \sqrt{2 + \sqrt{2(1 + \cos(4 \theta) } } \\ \\ \sf \implies \: \sqrt{2 + \sqrt{2(2 { cos }^{2} 2 \theta) } } \\ \\ \sf \implies \: \sqrt{2 + \sqrt{4 {cos}^{2}2 \theta } }$

Here,

$\sf \ \sqrt{4 {cos}^{2} 2 \theta} = 2 \cos(2 \theta) \\ \\ \sf \implies \: \sqrt{2 + \sqrt{4 {cos}^{2}2 \theta } } \\ \\ \sf \implies \sqrt{2 - {2 {cos}^{2} \theta } } \\ \\ \sf \implies \sqrt{2(2 {sin}^{2} \theta)} \\ \\ \sf \implies \sqrt{4 {sin}^{2} \theta } \\ \\ \sf \implies2 \sin( \theta)$