Question

Show the poccess how'd it occurred.

3(sinθ+cosθ)-2(sin³θ+cos³θ)=(sinθ+cosθ)³​

Answer 1

Answer:

Cos 3θ + i Sin3θ

Step-by-step explanation:

To prove --->

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(Cosθ + i Sinθ )³ = Cos3θ + i Sin3θ

Proof --->We have an identity

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(a + b )³ = a³ + b³ + 3 a b ( a + b )

Now

LHS =

( Cosθ + i Sinθ )³ =( Cosθ)³ + (i Sinθ )³ +3

Cosθ (i Sinθ ) ( Cosθ + i Sinθ )

= Cos³θ + i³ Sin³θ +3i Cos²θ Sinθ+3i²

Sin²θ Cosθ

=Cos³θ + i² i Sin³θ +3i Cos²θ Sinθ +3(-1)

Sin²θ Cosθ

=Cos³θ - 3 Sin²θ Cosθ + (-1) i Sin³θ +3 i

Cos²θ Sinθ

=Cos³θ -3(1 - Cos²θ ) Cosθ -i Sin³θ +3i

Cos²θ Sinθ

=Cos³θ -3Cosθ +3 Cos³θ - i (Sin³θ -3

Cos²θ Sinθ)

=4Cos³θ-3Cosθ - i{Sin³θ -3 (1-Sin²θ)Sinθ}

= Cos3θ - i (Sin³θ -3 Sinθ + 3 Sin³θ)

= Cos3θ - i (4 Sin³θ - 3Sinθ)

= Cos3θ + i (3 Sinθ - 4 Sin³θ)

=Cos 3θ + i Sin 3θ = RHS