Question

$\large\bold{\underline{\underline{Question:-}}}$


$\sf{If\:\dfrac{\cos\:x}{a_1}=\dfrac{\cos\:2x}{a_2}=\dfrac{\cos\:3x}{a_3}}$$\sf{Prove\: that\:\:\sin^2\dfrac{x}{2}=\dfrac{2a_2-a_1-a_3}{4a_2}}$

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Answer 1

Answer:

RHS=2a2−a1−a34a2. =14(2−a1a2−a3a2). =14(2−cosxcos2x−cos3x cos2x). =14(2−cosx+cos3xcos2x). =14(2−2cos2xcosxcos2x). =24(1−cosx). =12×2sin2(x2).

Answer 2

$\\\;\underbrace{\underline{\sf{Umderstanding\;the\;Concept\;:-}}}$

Here the Concept of Trigonometry has been used. We see we are given relation and we need to prove another given statement using the first one. Firstly we can simplify the equation to be proved and then apply the values we have from the first equation in it to find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{2\:\sin^{2}\:\dfrac{x}{2}\;=\;(1\;-\;\cos\:x)}}}$

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★ To Prove :-

$\\\;:\mapsto\;\;\bf{\blue{\sin^{2}\:\dfrac{x}{2}\;=\;\dfrac{2a_{2}\;-\;a_{1}\;-\;a_{3}}{4a_{2}}}}$

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★ Solution :-

Given,

$\\\;\odot\;\;\rm{\orange{\dfrac{\cos\:x}{a_{1}}\;=\;\dfrac{\cos\:2x}{a_{2}}\;=\;\dfrac{\cos\:3x}{a_{3}}}}$

From here by cross multiplication, we can derive that,

$\\\;\sf{\rightarrow\;\;\dfrac{a_{1}}{a_{2}}\;=\;\red{\dfrac{\cos\:x}{\cos\:2x}}}$

And,

$\\\;\sf{\rightarrow\;\;\dfrac{a_{3}}{a_{2}}\;=\;\red{\dfrac{\cos\:3x}{\cos\:2x}}}$

Now firstly, let's simplify the equation which is to be proved. Then,

$\\\;\sf{\green{:\rightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{2a_{2}\;-\;a_{1}\;-\;a_{3}}{4a_{2}}}}}$

By separating the numerator with denominator, this can be written as,

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\bigg(\dfrac{2a_{2}}{4a_{2}}\bigg)\;-\;\bigg(\dfrac{a_{1}}{4a_{2}}\bigg)\;-\;\bigg(\dfrac{a_{3}}{4a_{2}}\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\bigg(\dfrac{2}{4}\bigg)\;-\;\bigg(\dfrac{a_{1}}{4a_{2}}\bigg)\;-\;\bigg(\dfrac{a_{3}}{4a_{2}}\bigg)}}$

Now taking the denominator that is ¼ as common, we get

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{1}{4}\:\bigg(2\;-\;\dfrac{a_{1}}{a_{2}}\;-\;\dfrac{a_{3}}{a_{2}}\bigg)}}$

Now here applying the value which we got earlier, we get

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{1}{4}\:\bigg(2\;-\;\dfrac{\cos\:x}{\cos\:2x}\;-\;\dfrac{\cos\:3x}{\cos\:2x}\bigg)}}$

Taking - sign in common,

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{1}{4}\:\bigg(2\;-\;\bigg[\dfrac{\cos\:x}{\cos\:2x}\;+\;\dfrac{\cos\:3x}{\cos\:2x}\bigg]\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{1}{4}\:\bigg(2\;-\;\bigg[\dfrac{\cos\:x\;+\;\cos\:3x}{\cos\:2x}\bigg]\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{1}{4}\:\bigg(2\;-\;\bigg[\dfrac{2\cos\:2x\:\cos\:x}{\cos\:2x}\bigg]\bigg)}}$

Cancelling cos 2x , we get

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{1}{4}\:\bigg(2\;-\;\bigg[2\cos\:x\bigg]\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{1}{4}\:\bigg(2\;-\;2\cos\:x\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{1}{4}\:\bigg(2(1\;-\;\cos\:x)\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{2}{4}\:(1\;-\;\cos\:x)}}$

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{1}{2}\:(1\;-\;\cos\:x)}}$

We know that,

$\\\;\odot\;\;\sf{\pink{2\:\sin^{2}\:\dfrac{x}{2}\;=\;(1\;-\;\cos\:x)}}$

Now using this value and applying in equation,

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\dfrac{1}{2}\;\times\;(2\:sin^{2}\;\dfrac{x}{2})}}$

Cancelling 2, we get

$\\\;\sf{:\Longrightarrow\;\;\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\bigg(\sin^{2}\;\dfrac{x}{2}\bigg)}}$

$\\\;\sf{:\Longrightarrow\;\;\purple{\sin^{2}\:\dfrac{x}{2}\;=\;\bf{\sin^{2}\;\dfrac{x}{2}}}}$

Clearly LHS = RHS.

Hence, Verified .

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★ More to know :-

$\\\;\tt{\leadsto\;\;\sin^{2}\theta\;=\;1\;-\;\cos^{2}\theta}$

$\\\;\tt{\leadsto\;\;\sec^{2}\theta\;=\;1\;-\;\cot^{2}\theta}$

$\\\;\tt{\leadsto\;\;cosec^{2}\theta\;=\;1\;-\;\tan^{2}\theta}$

$\\\;\tt{\leadsto\;\;cosec\theta\;=\;\dfrac{1}{\sin\theta}}$

$\\\;\tt{\leadsto\;\;\sec\theta\;=\;\dfrac{1}{\cos\theta}}$

$\\\;\tt{\leadsto\;\;\cot\theta\;=\;\dfrac{1}{\tan\theta}}$