Question

Prove That
sin²6x - sin²4x = sin2x sin10x​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Trigonometric Identity has been used. We see that we are given a equation and we have to prove LHS = RHS. Firstly we shall use a algebraic identity to simplify this equation so that Trigonometric Identity can be applied. Then we shall apply the identity in the simplified equation to prove the given expression.

Let's do it !!

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★ Identities Used :-

$\\\;\boxed{\sf{\pink{1.)\;\;a^{2}\;-\;b^{2}\;=\;\bf{(a\;+\;b)(a\;-\;b)}}}}$

$\\\;\boxed{\sf{\pink{2.)\;\;\sin\:A\;+\;\sin\:B\;=\;\bf{2\sin\:\bigg(\dfrac{A\:+\:B}{2}\bigg)\cos\bigg(\dfrac{A\:-\:B}{2}\bigg)}}}}$

$\\\;\boxed{\sf{\pink{3.)\;\;\sin\:A\;-\;\sin\:B\;=\;\bf{2\cos\:\bigg(\dfrac{A\:+\:B}{2}\bigg)\sin\bigg(\dfrac{A\:-\:B}{2}\bigg)}}}}$

$\\\;\boxed{\sf{\pink{4.)\;\;\sin\:2\theta\;=\;\bf{2\sin\:\theta\:\cos\:\theta}}}}$

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★ To Prove :-

$\\\;\bf{\mapsto\;\;\blue{\sin^{2}\:6x\;-\;\sin^{2}\:4x\;=\;\sin\:2x\:\sin\:10x}}$

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Given,

$\\\;\bf{\odot\;\;L.H.S.\;=\;\sin^{2}\:6x\;-\;\sin^{2}\:4x}$

$\\\;\bf{\odot\;\;R.H.S.\;=\;\sin\:2x\:\sin\:10x}$

Now we shall simplify LHS separately to get the value of RHS.

Then,

$\\\;\sf{\rightarrow\;\;L.H.S.\;=\;\bf{\sin^{2}\:6x\;-\;\sin^{2}\:4x}}$

We know that,

$\\\;\tt{\gray{1.)\;\;a^{2}\;-\;b^{2}\;=\;(a\;+\;b)(a\;-\;b)}}$

• Here a² = sin² 6x

• And b² = sin² 4x

Now using the first identity, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{(\sin\:6x\;+\;\sin\:4x)(\sin\:6x\;-\;\sin\:4x)}}$

We know that,

$\\\;\tt{\gray{2.)\;\;\sin\:A\;+\;\sin\:B\;=\;2\sin\:\bigg(\dfrac{A\:+\:B}{2}\bigg)\cos\bigg(\dfrac{A\:-\:B}{2}\bigg)}}$

• Here A = 6x

• And B = 4x

Now using the second identity, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\bigg[2\sin\:\bigg(\dfrac{6x\:+\:4x}{2}\bigg)\cos\bigg(\dfrac{6x\:-\;4x}{2}\bigg)\bigg](\sin\:6x\;-\;\sin\:4x)}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\bigg[2\sin\:\bigg(\dfrac{10x}{2}\bigg)\cos\bigg(\dfrac{2x}{2}\bigg)\bigg](\sin\:6x\;-\;\sin\:4x)}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{\bigg[2\sin\:(5x)\cos\:(x)\bigg](\sin\:6x\;-\;\sin\:4x)}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{(2\sin\:5x\:\cos\:x)\bigg[(\sin\:6x\;-\;\sin\:4x)\bigg]}}$

We know that,

$\\\;\tt{\gray{3.)\;\;\sin\:A\;-\;\sin\:B\;=\;2\cos\:\bigg(\dfrac{A\:+\:B}{2}\bigg)\sin\bigg(\dfrac{A\:-\:B}{2}\bigg)}}$

• Here A = 6x

• And B = 4x

Now using the third identity, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{(2\sin\:5x\:\cos\:x)\bigg[2\cos\:\bigg(\dfrac{6x\:+\:4x}{2}\bigg)\sin\bigg(\dfrac{6x\:-\:4x}{2}\bigg)\bigg]}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{(2\sin\:5x\:\cos\:x)\bigg[2\cos\:\bigg(\dfrac{10x}{2}\bigg)\sin\bigg(\dfrac{2x}{2}\bigg)\bigg]}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{(2\sin\:5x\:\cos\:x)\bigg[2\cos\:(5x)\:\sin\:(x)\bigg]}}$

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{(2\sin\:5x\:\cos\:x)(2\cos\:5x\:\sin\:x)}}$

Now changing the position of 5, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{(2\sin\:5x\:\cos\:5x)(2\cos\:x\:\sin\:x)}}$

We know that,

$\\\;\tt{\gray{4.)\;\;\sin\:2\theta\;=\;2\sin\:\theta\:\cos\:\theta}}$

• Here θ = x

Now using fourth identity, we get

$\\\;\sf{\Longrightarrow\;\;L.H.S.\;=\;\bf{(\sin\:10x)(\sin\:2x)}}$

From this get,

$\\\;\bf{\red{\Longrightarrow\;\;L.H.S.\;=\;\bf{\sin\:10x\:\sin\:2x}}}$

$\\\;\bf{\orange{\Longrightarrow\;\;R.H.S.\;=\;\bf{\sin\:10x\:\sin\:2x}}}$

Clearly, LHS = RHS

$\\\;\bf{\purple{\Longrightarrow\;\;L.H.S.\;=\;R.H.S.\;=\;\bf{\sin\:10x\:\sin\:2x}}}$

$\\\qquad\qquad\underline{\boxed{\tt{Hence,\;\;Proved}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;\sin\:3x\;=\;3\sin\:x\;-\;4\sin^{3}\:x}$

$\\\;\sf{\leadsto\;\;\cos\:3x\;=\;4\cos^{3}\:x\;-\;3\cos\:x}$

$\\\;\sf{\leadsto\;\;\cos(2\pi\;-\;x)\;=\;\cos\:x}$

$\\\;\sf{\leadsto\;\;\sin(2\pi\;-\;x)\;=\;-\:\sin\:x}$

$\\\;\sf{\leadsto\;\;\sin(-\:x)\;=\;-\:sin\:x}$

$\\\;\sf{\leadsto\;\;\cos(-\:x)\;=\;\cos\:x}$

Answer 2

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

$\bf \:Prove \: that : sin²6x - sin²4x = sin2x sin10x$

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$\large\underline\blue{\bold{Formula \: used:- }}$

$\bf \: {sin}^{2} x - {sin}^{2} y = sin(x + y)sin(x - y)$

☆Proof of this result :-

☆Consider RHS :-

$\sf \: ⟼sin(x + y)sin(x - y)$

$\sf \: =( sinxcosy + sinycosx)(sinxcosy - sinycosx)$

$\sf \: = {sin}^{2} x {cos}^{2} y - {cos}^{2} x {sin}^{2} y$

$\sf \: = {sin}^{2} x(1 - {sin}^{2} y) - {sin}^{2}y (1 - {sin}^{2} x)$

$\sf \: = {sin}^{2} x - {sin}^{2} x {sin}^{2}y - {sin}^{2} y + {sin}^{2} x {sin}^{2} y$

$\large \blue{\sf \: = {sin}^{2} x - {sin}^{2}y }$

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$\large\underline\purple{\bold{Solution :- }}$

$\bf \:sin²6x - sin²4x$

☆Using identity, we get

$\bf \: = sin(6x + 4x)sin(6x - 4x)$

$\bf \: = sin10x \: sin2x$

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$\large \purple{\bf \: ⟼ Explore \: more } ✍$

Trigonometry Formulas

• sin(−θ) = −sin θ
• cos(−θ) = cos θ
• tan(−θ) = −tan θ
• cosec(−θ) = −cosecθ
• sec(−θ) = sec θ
• cot(−θ) = −cot θ

Product to Sum Formulas

• sin x sin y = 1/2 [cos(x–y) − cos(x+y)]
• cos x cos y = 1/2[cos(x–y) + cos(x+y)]
• sin x cos y = 1/2[sin(x+y) + sin(x−y)]
• cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

• sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]
• sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]
• cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]
• cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

• cos (A + B) = cos A cos B – sin A sin B
• cos (A – B) = cos A cos B + sin A sin B
• sin (A+B) = sin A cos B + cos A sin B
• sin (A -B) = sin A cos B – cos A sin B
• tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]
• tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]
• cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]
• cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]
• cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A
• sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

• sin2A = 2sinA cosA = [2tan A /(1+tan²A)]
• cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]
• tan 2A = (2 tan A)/(1-tan²A)

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