Q-1-1-cos-2-2A-2-cos-4-A-sin-4-A-2-sin-2-A-sin-2-120-0-A-sin-2-120-0-A-3-2- Tinku Tara June 3, 2023 Trigonometry 0 Comments FacebookTweetPin Question Number 8401 by rhm last updated on 10/Oct/16 Q.11+cos22A=2(cos4A+sin4A)2.sin2A+sin2(1200+A)+sin2(1200−A)=32 Answered by ridwan balatif last updated on 10/Oct/16 Q.11+cos22A=2(cos4A+sin4A)Missing \left or extra \rightMissing \left or extra \right=2((cos2A+sin2A)2−2sin2Acos2A)=2(1−(2sinAcosA).22sinAcosA)=2(1−sin22A2)=2−sin22A=1+1−sin22A=1+cos22A∴1+cos2A=2(cos4A+sin4A)(PROVE) Answered by ridwan balatif last updated on 10/Oct/16 2.sin2A+(sin(120o+A))2+(sin(120o−A))2sin2A+(sin120ocosA+sinAcos120o)2+(sin1200cosA−sinAcos120o)2sin2A+(123cosA−12sinA)2+(123cosA+12sinA)2sin2A+{(123cosA−12sinA)+(123cosA+12sinA)}2−2(123cosA−12sinA)(123cosA+12sinA)sin2A+(3cosA)2−2(34cos2A−14sin2A)32sin2A+32cos2A32(sin2A+cos2A)32∴sin2A+sin2(120o+A)+sin2(120o−A)=32 Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: prove-3-2-9-4-2-2-2-Next Next post: Q-cos-2-66-0-sin-2-6-0-cos-2-48-0-sin-2-12-0-1-16- Leave a Reply Cancel replyYour email address will not be published.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.