Question

Hola Brainlians. Here's a question to check your Trigonometric skill. ICSE Board : Inverse Trigonometry, Class 10. Give answer fast and no spam.

Proof the given identity.

sin^{-1} (2x√(1 - x²)) = 2 sin^{-1} x , -1/√2 ≤ x ≤ 1/2

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Answer 1

★ Concept :-

Here the concept of Inverse Trigonometry has been used. We see that we are given an equation to be proved. The easiest method to do it is using substitution method. Firstly we can take any value dependent on the main function of this equation. Then we can apply that in the main equation and thus by applying other trignometric identities we can prove the answer.

Let's do it !!

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★ Solution :-

Let sin¯¹ x = y

From this we can get,

>> x = sin y

Now we shall apply this value in the equation to be proved.

From the things to be proved, we have

$\;\;\sf{\odot\;\;\green{L.H.S.\;=\;\sin^{-1}\bigg(2x\sqrt{1\:-\:x^{2}}\bigg)}}$

And,

$\;\;\sf{\odot\;\;\blue{R.H.S.\;=\;2\sin^{-1}x}}$

Also we are given some general information that,

$\;\;\bf{\odot\;\;\orange{-\dfrac{1}{\sqrt{2}}\:\leq\:x\:\leq\:\dfrac{1}{2}}}$

Now let's start applying the value we assumed in the equation.

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\sin^{-1}\bigg(2x\sqrt{1\:-\:x^{2}}\bigg)}$

By applying the value of x, we get

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\sin^{-1}\bigg(2(\sin y)\sqrt{1\:-\:(\sin y)^{2}}\bigg)}$

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\sin^{-1}\bigg(2(\sin y)\sqrt{1\:-\:\sin^{2}y}\bigg)}$

We know that : cos² A = 1 - sin² A

By applying this, we get

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\sin^{-1}\bigg(2(\sin y)\sqrt{\cos^{2} y}\bigg)}$

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\sin^{-1}\bigg(2(\sin y)(\cos y)\bigg)}$

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\sin^{-1}\bigg(2\sin y\cos y\bigg)}$

We know that : sin 2A = 2 sin A cos A

By applying this, we get

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\sin^{-1}\bigg(\sin 2y\bigg)}$

Now sin¯¹ and sin will cancel each other giving 1. We get,

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;1\:\times\:2y}$

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;2y}$

Now let's apply the value of y. We get,

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;2(\sin^{-1}x)}$

$\;\;\bf{\rightarrow\;\;\red{L.H.S.\;=\;2\sin^{-1}x}}$

This is equal to R.H.S. By applying that, we get,

$\;\;\bf{\rightarrow\;\;\purple{L.H.S.\;=\;R.H.S.\;=\;2\sin^{-1}x}}$

• Hence, proved.

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★ More to know :-

• sin² A + cos² A = 1

• cosec² A = 1 + cot² A

• sec² A = 1 + tan² A

• cos 2A = 2 cos² A - 1

• sin 3A = 3 sin A - 4 sin³ A

• cos 3A = 4 cos³ A - 3 cos A

Answer 2

Answer:

Answer in the attachment.