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  • 11-10-2020
  • Mathematics
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(asin θ-bcos θ)²=?​​

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cromulent
cromulent cromulent
  • 11-10-2020

[tex](a\sin(\theta) - b\cos(\theta))^2\\= a^2\sin^2(\theta) - 2ab\sin(\theta)\cos(\theta) + b^2\cos^2(\theta)\\= a^2\sin^2(\theta) - ab\sin(2\theta) + b^2\cos^2(\theta)[/tex]

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