मराठी

Prove that: cos 3 2 x + 3 cos 2 x = 4 ( cos 6 x − sin 6 x )

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प्रश्न

Prove that: \[\cos^3 2x + 3 \cos 2x = 4\left( \cos^6 x - \sin^6 x \right)\]

संख्यात्मक
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उत्तर

Let us consider the RHS

4(cos6 x – sin6 x)

Now, upon expansion we get,

4(cos6 x – sin6 x) = 4[(cos2 x)3 – (sin2 x)3]

= 4(cos2 x – sin2 x) (cos4 x + sin4 x + cos2 x sin2 x)

On using the formula,

a3 – b3 = (a - b) (a2 + b2 + ab)

= 4 cos 2x (cos4 x + sin4 x + cos2 x sin2 x + cos2 x sin2 x – cos2 x sin

As we know, cos 2x = cos2 x – sin2 x

Therefore,

= 4 cos 2x (cos4 x + sin4 x + 2 cos2 x sin2 x – cos2 x sin2 x)

= 4 cos 2x [(cos2 x)2 + (sin2 x)2 + 2 cos2 x sin2 x – cos2 x sin2 x]

As we know, a2 + b2 + 2ab = (a + b)2

= 4 cos 2x [(1)2 – 1/4 (4 cos2 x sin2 x)]

= 4 cos 2x [(1)2 – 1/4 (2 cos x sin x)2]

Again as we know, sin 2x = 2sin x cos x

= 4 cos 2x [(12) – 1/4 (sin 2x)2]

= 4 cos 2x (1 – 1/4 sin2 2x)

As as we know, sin2 x = 1 – cos2 x

= 4 cos 2x [1 – 1/4 (1 – cos2 2x)]

= 4 cos 2x [1 – 1/4 + 1/4 cos2 2x]

= 4 cos 2x [3/4 + 1/4 cos2 2x]

= 4 (3/4 cos 2x + 1/4 cos3 2x)

= 3 cos 2x + cos3 2x

= cos3 2x + 3 cos 2x

= LHS

Thus proved. 

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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पाठ 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [पृष्ठ २८]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 14 | पृष्ठ २८

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