cos 3A = 4 cos³A – 3cosA
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Answered by
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Cos 3A = Cos ( 2A+ A)
= Cos 2A Cos A - Sin 2A Sin A
= ( 2 Cos²A -1) Cos A - 2 Sin²A Cos A
= 2 Cos³A -Cos A - 2 Cos A( 1- Cos²A)
= 4 Cos³ A- 3 Cos A
HENCE PROVED
= Cos 2A Cos A - Sin 2A Sin A
= ( 2 Cos²A -1) Cos A - 2 Sin²A Cos A
= 2 Cos³A -Cos A - 2 Cos A( 1- Cos²A)
= 4 Cos³ A- 3 Cos A
HENCE PROVED
ruchisr123:
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Answered by
10
Cos 3A = Cos ( 2A+ A)
= Cos 2A Cos A - Sin 2A Sin A
= ( 2 Cos²A -1) Cos A - 2 Sin²A Cos A
= 2 Cos³A -Cos A - 2 Cos A( 1- Cos²A)
= 4 Cos³ A- 3 Cos A
∴ cos 3A= 4 cos³A- 3cosA
= Cos 2A Cos A - Sin 2A Sin A
= ( 2 Cos²A -1) Cos A - 2 Sin²A Cos A
= 2 Cos³A -Cos A - 2 Cos A( 1- Cos²A)
= 4 Cos³ A- 3 Cos A
∴ cos 3A= 4 cos³A- 3cosA
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