Please help me with this problem.. i couldnt even get anything nearer to it.
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To Prove : cosֿ¹ x = 2 · sinֿ¹ √[ ( 1 - x ) / 2 ].
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On the RHS,
let : x = cos 2Φ.
∴ 2Φ = cosֿ¹ x ........................................… (1)
∴√[ (1-x) / 2] = √[ ( 1 - cos 2Φ ) / 2 ]
. . . . . . . . . . . = √[ ( 2 sin² Φ ) / 2 ]
. . . . . . . . . . . = sin Φ.
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∴ RHS
= 2 · sinֿ¹ √[ (1-x) / 2 ]
= 2 · sinֿ¹ ( sin Φ )
= 2 · Φ
= ( 2Φ )
= cosֿ¹ x ............... from (1)
= LHS ........................................… Q.E.D.
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Happy To Help !
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On the RHS,
let : x = cos 2Φ.
∴ 2Φ = cosֿ¹ x ........................................… (1)
∴√[ (1-x) / 2] = √[ ( 1 - cos 2Φ ) / 2 ]
. . . . . . . . . . . = √[ ( 2 sin² Φ ) / 2 ]
. . . . . . . . . . . = sin Φ.
_____________________________
∴ RHS
= 2 · sinֿ¹ √[ (1-x) / 2 ]
= 2 · sinֿ¹ ( sin Φ )
= 2 · Φ
= ( 2Φ )
= cosֿ¹ x ............... from (1)
= LHS ........................................… Q.E.D.
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Happy To Help !
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