Prove-that-sin-5A-5-cos-4-A-sin-A-10-cos-2-A-sin-3-A-sin-5-A- Tinku Tara June 4, 2023 Trigonometry 0 Comments FacebookTweetPin Question Number 17866 by Tinkutara last updated on 11/Jul/17 Provethatsin5A=5cos4AsinA−10cos2Asin3A+sin5A Answered by alex041103 last updated on 13/Jul/17 sin5A=sin(2A+3A)Weusesin(A+B)=sinAcosB+sinBcosAsin2A=2sinAcosAcos2A=cos2A−sin2Asin3A=3sinA−4sin3Acos3A=4cos3A−3cosAAndwehavesin5A=8sinAcos4A−3sinAcos2A−3sin3A−4sin3Acos2A+4sin5A==5cos4AsinA−10cos2Asin3A+sin5A+3sinA(sin4A+2sin2Acos2A+cos4A−sin2A−cos2A)=5cos4AsinA−10cos2Asin3A+sin5A+3sinA[(sin2A+cos2A)2−(sin2A+cos2A)]=5cos4AsinA−10cos2Asin3A+sin5A+3sinA(12−1)==5cos4AsinA−10cos2Asin3A+sin5A+3sinA(0)==5cos4AsinA−10cos2Asin3A+sin5A⇒sin5A=5cos4AsinA−10cos2Asin3A+sin5A Commented by Tinkutara last updated on 13/Jul/17 ThanksSir! Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: B-3-2B-2-4B-8-y-0-solve-the-differencial-equation-Next Next post: prove-that-2-3-2n-2-3-2n-is-an-even-integer-and-that-2-3-2n-2-3-2n-w-3-for-some-integers-w-for-all-integer-n-1- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![sin5A=sin(2A+3A) We use sin(A+B)=sinAcosB+sinBcosA sin2A=2sinAcosA cos2A=cos^2 A−sin^2 A sin3A=3sinA−4sin^3 A cos3A=4cos^3 A−3cosA And we have sin5A=8sinAcos^4 A−3sinAcos^2 A −3sin^3 A−4sin^3 Acos^2 A+4sin^5 A= =5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A +3sinA(sin^4 A+2sin^2 Acos^2 A+cos^4 A−sin^2 A−cos^2 A) =5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A +3sinA[(sin^2 A+cos^2 A)^2 −(sin^2 A+cos^2 A)] =5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A +3sinA(1^2 −1)= =5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A +3sinA(0)= =5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A ⇒sin5A=5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A](https://www.tinkutara.com/question/Q17967.png)
