Pre Calculus

posted by .

Establish the following identity

sin 2x/sinx - cos2x/cosx = secx

  • Pre Calculus -

    Use sin 2x = 2sinxcosx and
    cos 2x = 2cos^2x - 1
    in the left side, it comes apart nicely

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Pre-Calc

    Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx + cosx/cosx)/(sinx/cos x - 1/cosx + cosx/cosx)= tanx + cosx (just working on the left side) ((sinx + 1 - cosx)/cosx)/((sinx …
  2. Pre-Calculus

    How can this identity be proved/verified?
  3. Pre-Calculus

    How do you verify: 1.) sinxcosx+sinx^3secx=tanx 2.) (secx+tanx)/(secx-tan)=(secx+tanx)^2 I tried starting from the left on both problems, but am stuck. like I said before, change everything into sines and cosines. for the first one: …
  4. drwls

    My previous question: Verify that (secx/sinx)*(cotx/cscx)=cscx is an identity. (secx/sinx)*(cotx/cscx) = (secx/cscx)(cotx/sinx) = (sinx/cosx)*cotx*(1/sinx) "The last steps should be obvious" Not to me. I can convert (sinx/cosx) to …
  5. Math - Pre- Clac

    Prove that each of these equations is an identity. A) (1 + sinx + cos x)/(1 + sinx + cosx)=(1 + cosx)/sinx B) (1 + sinx + cosx)/(1 - sinx + cosx)= (1 + sin x)/cosx Please and thankyou!
  6. Pre Calculus

    Establish the identity (cosx/2-sinx/2)^2=1-sin x
  7. Pre Calculus

    Having trouble verifying this identity... cscxtanx + secx= 2cosx This is what I've been trying (1/sinx)(sinx/cosx) + secx = 2cosx (1/cosx)+(1/cosx) = 2 cosx 2 secx = 2cosx that's what i ended up with, but i know it's now right. did …
  8. Trigonometry

    Prove the following trigonometric identities. please give a detailed answer because I don't understand this at all. a. sin(x)tan(x)=cos(x)/cot^2 (x) b. (1+tanx)^2=sec^2 (x)+2tan(x) c. 1/sin(x) + 1/cos(x) = (cosx+sinx)(secx)(cscx) d. …
  9. Precalculus/Trig

    I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx I don't know …
  10. Calculus 2 Trigonometric Substitution

    I'm working this problem: ∫ [1-tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx) ∫cosx-sinx(sinx/cosx) ∫cosx-∫sin^2(x)/cosx sinx-∫(1-cos^2(x))/cosx sinx-∫(1/cosx)-cosx sinx-∫secx-∫cosx …

More Similar Questions