Given \tan\alpha=2, evaluate \frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}

Aaden Lam

Aaden Lam

Answered question

2022-01-26

Given tanα=2, evaluate sin3α2cos3α+3cosα3sinα+2cosα

Answer & Explanation

ataill0k

ataill0k

Beginner2022-01-27Added 18 answers

Notice sinα=2cosα and cos2α=11+tan2α=15
so we have
sin3α2cos3α+3cosα3sinα+2cosα=8cos3α2cos3α+3cosα6cosα+2cosα
=6cos3α+3cosα8cosα=6cos2α+38=2140

Aiden Cooper

Aiden Cooper

Beginner2022-01-28Added 14 answers

sin3α2cos3α+3cosα3sinα+2cosα=tan3α2+3sec2α(3tanα+2)sec2α
where
sec2α=tan2α+1
Hence
2140

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