Question

Prove LHS = RHS

Cos9x - Cos5x / Sin17x -Sin3x = -Sin2x / cos10x

Answer 1

Heya !
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★Trigonometry ★
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$= > \frac{cos \: 9x - \: cos \: 5x}{sin \: 17x - sin3x} = \frac{ - sin2x}{cos \: 10x}$



→ Taking L.H.S ,


$= > \frac{ - 2 \: sin \: \frac{9x + 5x}{2} \: sin \: \frac{9x - 5x}{2} }{2 \: cos \: \frac{17x + 3x}{2} \: sin \: \frac{17x - 3x}{2} }$



→ Solving and combining the terms we have ,


$= > \frac{ - 2 \: sin \: 7x \: sin \: 2x}{2 \: cos \: 10x \: sin \: 7x}$

★ Now , 2 gets cancelled out on the numerator and denominator . Also , Sin 7x gets cancelled as a common factor . Thus we are left with ,


$= > \frac{ - sin \: 2x}{cos \: 10x}$


•°• L.H.S = R.H.S



Identities used
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$= > cos \: x \: - \: cos \: y = - sin \: \frac{x + y}{2} \: sin \: \frac{x - y}{2}$

$= > sin \: x \: - \: sin \: y = 2 \: cos \: \frac{x + y}{2} \: sin \: \frac{x - y}{2}$

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Answer 2

Step-by-step explanation:

Answer:

Step-by-step explanation:

cos A-cos B=-2 sin ((A+B)/2) sin ((A-B)/2)

cos9x-cos 5x=-2 sin((9x+5x)/2) sin((9x-5x)/2)=-2 sin 7x sin2x

sin A -sin B=2 cos ((A+B)/2) sin ((A-B)/2)

sin 17x-sin 3x=2 cos ((17x+3x)/2) sin ((17x-3x)/2)=2 cos 10x sin 7x

Now

cos9x-cos 5x/sin 17x-sin 3x = -2 sin 7x sin2x/2 cos 10x sin 7x

cancel sin7x and 2

Now it becomes

cos9x-cos 5x/sin 17x-sin 3x= - sin2x/cos10x