Answered

Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

Sin A - 2 sin cube A
---------------------------
2 cos cube A - cos A

Sagot :

[tex]\frac{sin\alpha-2sin^3\alpha}{2cos^3\alpha-cos\alpha}=\frac{sin\alpha(1-2sin^2\alpha)}{cos\alpha(2cos^2\alpha-1)}=\frac{sin\alpha(1-sin^2\alpha-sin^2\alpha)}{cos\alpha(cos^2\alpha+cos^2\alpha-1)}\\\\=\frac{sin\alpha(cos^2\alpha-sin^2\alpha)}{cos\alpha(cos^2\alpha-sin^2\alpha)}=\frac{sin\alpha}{cos\alpha}=tan\alpha\\\\\\sin^2\alpha+cos^2\alpha=1\to cos^2\alpha=1-sin^2\alpha\\sin^2\alpha+cos^2\alpha=1\to sin^2\alpha=1-cos^2\alpha\to-sin^2\alpha=cos^2\alpha-1[/tex]
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.