Question

If sin 27" - p, then the value of VI + sin 36° is​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The correct answer is  $\sqrt{1+sin 36}$° = $\sqrt{2-2p^{2} }$

Given:

p= sin 27°

To Find:

$\sqrt{1+sin36}$° = ?

Solution:

We have been given that p= sin 27°. To find $\sqrt{1+sin 36}$°, we need to make use of the following trigonometric identities

1. sin(90-A)= cosA

2. sin(A-B)= sinAcosB - cosAsinB

3. cos2A= 1-2$sin^{2}A$

Now,

$\sqrt{1+sin36}$° = $\sqrt{1+sin(90-54)}$°

⇒$\sqrt{1+sin36}$°=$\sqrt{1+sin90cos54-cos90sin54}$ ............................from (2)

Now, sin90° = 1 and cos90° = 0. Substituting this we get

$\sqrt{1+sin36}$° = $\sqrt{1+ cos54}$° = $\sqrt{1+ cos(2*27)$

⇒$\sqrt{1+sin36}$° = $\sqrt{1+ 1-2sin^{2}A$

⇒$\sqrt{1+sin36}$° = $\sqrt{2-2sin^{2}A$

⇒$\sqrt{1+sin36}$° = $\sqrt{2-2p^{2}$

Therefore $\sqrt{1+sin36}$° = $\sqrt{2-2p^{2}$

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