Question

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Answer 1

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Given:

m = acos³Ѳ + 3acosѲ.sin²Ѳ

n = asin³Ѳ +3acos²Ѳ.sinѲ

We have to prove that , [∛(m+n)]² + [∛(m-n)]² = 2[∛a]²

Step-by-step explanation:

Substituting the values in the above equation,

we get..,

=     [∛(acos³Ѳ + 3acosѲ.sin²Ѳ+ asin³Ѳ +3acos²Ѳ.sinѲ)]²                                                       . + . [∛(acos³Ѳ + 3acosѲ.sin²Ѳ- asin³Ѳ - 3acos²Ѳ.sinѲ )]²

= ∛[a(cos²Ѳ+sin²Ѳ) + 3acosѲ.sinѲ(sinѲ+cosѲ)]²                                          +  ∛[a(cos²Ѳ-sin²Ѳ) -  3acosѲ.sinѲ(sinѲ+cosѲ)]²

applying trigonometric identities and squaring finally cancelling ,

= ∛[2a]²

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Answer 2

Answer:

Step-by-step explanation: