Question

3cos⁻¹ x = cos⁻¹ (4x³ - 3x), x∈[1/2, 1 ] को सिद्ध कीजिये

Answer 1

Question :

To prove :

3cos⁻¹ x = cos⁻¹ (4x³ - 3x), x∈[1/2, 1 ]

Properties of inverse Function :

1)cos⁻¹(cos x)= x

2) sin⁻¹(sinx)=x

3)tan⁻¹(tanx)=x

Note : these properties are within the range of inverse functions .

Solution :

we have to prove that 3cos⁻¹ x = cos⁻¹ (4x³ - 3x)

___________________

Let x =cosθ

RHS = cos⁻¹ (4x³ - 3x)

$= \cos {}^{ - 1} (4cos{}^{3} \theta - 3cos \theta)$

we know that cos3x= 4cos³x- 3cosx

$= \cos {}^{ - 1} ( \cos( 3 \theta) )$

Now cos⁻¹(cosx)= x

$= 3 \theta$

since x =cosθ i.e cos⁻¹ x = θ

$= 3 \cos {}^{ - 1} (x)$

LHS =3cos⁻¹ x

⇒LHS = RHS

$\huge{\bold{Hence\: Proved}}$

Answer 2

Given   :     3cos⁻¹ x = cos⁻¹ (4x³ - 3x), x∈[1/2, 1 ]

To find :   सिद्ध कीजिए

Solution:

3cos⁻¹ x = cos⁻¹ (4x³ - 3x)

माना x = Cosα

LHS = 3Cos⁻¹ x

= 3cos⁻¹ (Cosα)

= 3α

RHS =

cos⁻¹ (4x³ - 3x)

= Cos⁻¹ ( 4Cosα³ - 3Cosα)

हमें ज्ञात है की   Cos3α = 4Cosα³ - 3Cosα

= Cos⁻¹ ( Cos3α)

= 3α

LHS = RHS = 3α

QED

इति सिद्धम

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