Math, asked by needhelpmods, 6 hours ago

Prove
cos9x-cos5x /sin17x-sin3x=-sin2x /cos10x
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Answers

Answered by sethrollins13
160

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

____________________

Answered by MяMαgıcıαη
168

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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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