Question

prove that sin3x=3sinx-4sin^3x​

Answer 1

Solution:

Given Equation:

=> sin 3x = 3 sin x - 4 sin³x

So, we will take LHS part,

=> sin 3x

By using identity, sin(A + B) = sin A cos B + cos A sin B

=> sin 2x. cos x + cos 2x. sin x

We know that,

[sin 2x = 2 sin x cos x]

[cos 2x = 1 - 2 sin²x]

So,

=> 2 sin x cos x. cos x + (1 - 2 sin²x). sin x        

=> 2 sin x cos x. cos x + sin x - 2 sin³x

=> 2 sin x cos²x + sin x - 2 sin³x

=> 2 sin x (1 - sin²x) + sin x - 2 sin³x  

=> 2 sin x - 2 sin³x + sin x - 2 sin³x

=> 3 sin x - 4 sin³x

Answer 2

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According to the given question:-

$\sf⟶(sin \: 3x = 3 \: sin x - 4 \: sin {}^{3} x)$

Taking LCM

$\sf⟶sin\:3x$

Identity using here

$\sf⟶sin(A + B) = sin \: A cos \: B + cos \: A \: sin \: B$

$\sf⟶sin \: 2x \: (cos \: x + cos \: 2x) \: sin \: x$

Given

$\sf⟶(sin \: 2x \: = \: 2 \: sin \: x \: cos \: x )\\ </p><p></p><p>\sf⟶(cos \: 2x \: = \: 1 - 2 \: sin {}^{2} x)$

Hence,

$\sf⟶2 \: sin \: x \: cos \: x \: cos \: x + (1 - 2 \: sin {}^{2} x) sin \: x$

$\sf⟶2 sin \: x \: cos \: x \: cos \: x + sin \: x - 2 sin \: {}^{3} x$

$\sf⟶(2 sin \: x \: cos {}^{2} x + sin \: x - 2 {}^{3} sin \: x)$

$\sf⟶2 sin x (1 - sin {}^{2} x) + sin x - 2 {}^{3} sinx$

$\sf⟶2 sin \: x - 2 sinx{3}^{2} + sin x - 2 {3}^{2} sin \: x$

$\sf⟶(3 sin \: x - 4 sin {}^{3} x)$