Prove that: (cosx)^3*(sinx)^2=1/16*(2cosx-cos3x-cos5x)
- print Print
- list Cite
Expert Answers
beckden
| Certified Educator
calendarEducator since 2011
write562 answers
starTop subjects are Math, Science, and Business
`cos2x = cos^2x - sin^2x `
`sin2x = 2sinxcosx`
`cos3x = cosxcos2x - sinxsin2x = cosx(cos^2x-sin^2x) - sinx(2sinxcosx) `
`cos3x = cos^3x - cosxsin^2x - 2sin^2xcosx = cos^3x - 3sin^2xcosx `
`cos3x = cos^3x - 3(1-cos^2x)cosx = cos^3x - 3cosx + 3cos^3x `
`cos3x = 4cos^3x - 3cosx `
`sin3x = sin2xcosx + cos2xsinx = 2cosxsinxcosx + (2cos^2x - 1)sinx `
`sin3x = 2cos^2xsinx +...
(The entire section contains 242 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- Solve the equation (cosx+cos3x+cos5x)/(sinx+sin3x+sin5x)=square root of 3.
- 1 Educator Answer
- Prove the identity `{1-sinx}/cosx=cosx/{1+sinx}`
- 1 Educator Answer
- Prove ((sinx+cosx)/(sinx-cosx))+((sinx-cosx)/(sinx+cosx)) = (2sec^2 x/tan^2 x)-1
- 1 Educator Answer
- How to prove trig identity (sin3x/sinx)-(cos3x/cosx)=2
- 1 Educator Answer
- Prove the identity sinx/2=squareroot(1-cosx)/2.
- 1 Educator Answer