Question

Prove
cos9x-cos5x /sin17x-sin3x=-sin2x /cos10x
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Answer 1

Given :

• cos 9x - cos 5x / sin 17x - sin 3x

To Prove :

• cos 9x - cos 5x / sin 17x - sin 3x = -sin 2x / cos 10 x .

Solution :

Using Identities :

• $cos\:x-cos\:y=-2sin\dfrac{x+y}{2}\:sin\dfrac{x-y}{2}$

• $sin\:x-sin\:y=2cos\dfrac{x+y}{2}\:sin\dfrac{x-y}{2}$

L.H.S :

$\longmapsto\tt{\dfrac{-\bigg(2\:sin\dfrac{9x+5x}{2}\:sin\dfrac{9x-5x}{2}\bigg)}{2\:cos\dfrac{17x+3x}{2}\:sin\dfrac{17x-3x}{2}}}$

$\longmapsto\tt{\dfrac{-\bigg(2\:sin\dfrac{{\cancel{14x}}}{{\cancel{2}}}\:sin\dfrac{{\cancel{4x}}}{{\cancel{2}}}\bigg)}{2\:cos\dfrac{{\cancel{20x}}}{{\cancel{2}}}\:sin\dfrac{{\cancel{14x}}}{{\cancel{2}}}}}$

$\longmapsto\tt{\dfrac{-{\cancel{2}}\:{\cancel{sin\:7x}}\:sin\:2x}{{\cancel{2}}\:cos\:10x\:{\cancel{sin\:7x}}}}$

$\longmapsto\tt\bf{\dfrac{-sin\:2x}{cos\:10x}}$

HENCE PROVED !

____________________

Some more Trigonometric Identities :

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

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Answer 2

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$\underline{\underline{\sf{\red{Given\::-}}}}$

• $\sf \dfrac{cos9x - cos5x}{sin17x - sin3x}$

$\underline{\underline{\sf{\red{To\:Prove\::-}}}}$

• $\sf \dfrac{cos9x - cos5x}{sin17x - sin3x} = \dfrac{-sin2x}{cos10x}$

$\underline{\underline{\sf{\red{Proof\::-}}}}$

⌬ Identities that we will use :-

• $\sf cos\:A - cos\:B = -2sin\bigg\lgroup \dfrac{A + B}{2} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{A - B}{2} \bigg\rgroup$

• $\sf sin\:A - sin\:B = 2cos\bigg\lgroup \dfrac{A + B}{2} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{A - B}{2} \bigg\rgroup$

Now,

$\longrightarrow\:\sf L.H.S = \dfrac{cos9x - cos5x}{sin17x - sin3x}$

$\longrightarrow\:\sf L.H.S = \dfrac{-2sin\bigg\lgroup \dfrac{9x + 5x}{2} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{9x - 5x}{2} \bigg\rgroup}{2cos\bigg\lgroup \dfrac{17x + 3x}{2} \bigg\rgroup\:.\: sin \bigg\lgroup \dfrac{17x - 3x}{2} \bigg\rgroup}$

$\longrightarrow\:\sf L.H.S = \dfrac{-2sin\bigg\lgroup \dfrac{14x}{2} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{4x}{2} \bigg\rgroup}{2cos\bigg\lgroup \dfrac{20x}{2} \bigg\rgroup\:.\: sin \bigg\lgroup \dfrac{14x}{2} \bigg\rgroup}$

$\longrightarrow\:\sf L.H.S = \dfrac{-2sin\bigg\lgroup \dfrac{\cancel{14}x}{\cancel{2}} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{\cancel{4}x}{\cancel{2}} \bigg\rgroup}{2cos\bigg\lgroup \dfrac{\cancel{20}x}{\cancel{2}} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{\cancel{14}x}{\cancel{2}} \bigg\rgroup}\:(Cancelling)$

$\longrightarrow\:\sf L.H.S = \dfrac{-2sin7x\:.\:sin2x}{2cos10x\:.\:sin7x}$

$\longrightarrow\:\sf L.H.S = \dfrac{-\cancel{2}\cancel{sin7x}\:.\:sin2x}{\cancel{2}cos10x\:.\:\cancel{sin7x}}\:(Cancelling)$

$\longrightarrow\:\sf L.H.S = \dfrac{-sin2x}{cos10x}$

$\pink{\bigstar}\:\underline{\underline{\bf{\blue{L.H.S = R H.S}}}}$

$\:\:\qquad\qquad\therefore\:\underline{\underline{\sf{\red{Hence,\:Proved!}}}}$

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