Question

निम्नलिखित को सिद्ध कीजिए: $\cos 4x = 1 - 8\sin^{2}x\,cos^{2}x$

Answer 1

Prove that : Cos 4x = 1 - 8 Sin²x.Cos²x


From RHS ;

↪1 - 8 Sin²x.Cos²x

↪1 - 2 ( 2Sinx.Cosx )²

↪1 - 2 (Sin2x)² [ •°• Sin 2x = 2Sin x . Cos x ]

↪1 - 2 Sin²2x

↪Cos 4x [ •°• Cos 2x = 1 - 2 Sin²x ] , or [ Cos 4x = 1 - 2 Sin²2x]

⬆Note : Here, We have replaced x = 2x ] ⬆



Trigonometric Function Formula Used !

↪ Sin 2x = 2Sin x . Cos x

↪ Cos 2x = 1 - 2 Sin²x

↪ Cos 4x = 1 - 2 Sin²2x

Answer 2

Step-by-step Proof :

Given,

L. H. S

=cos4 x

= 1 - 2 sin²2x

= 1 - 2 ( sin2x)²

= 1 - 2 ( 2sinx. cosx)²

= 1 - 2 ( 4 sin²x . cos²x)

= 1 - 8 sin²x cos²x

= RHS

We used the Trigonometry ratios of multiples,

1) sin2p = 2sinp. cosp

2) cos2p = 1 - 2sin²p = 2cos²p - 1

हम सिद्ध किया की $\cos 4x = 1 - 8\sin^{2}x\,cos^{2}x$