A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

Not sure the right approach to solve the identity: tan3x - tanx = 2sinxsecx

  • This Question is Closed
  1. Owlfred
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry i meant tan3x - tanx = 2sinxsec3x

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this is a tricky - you might try converting to functions of sin and cos

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i've spent a lot of time trying diff techniques, splitting 3x in sinx and cosx components but am getting no where fast!

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah - i've got a headache trying this one but I've got to go out now. I'll look at it later. Maybe someone else can help meantime. I'd like to see the proof of this one. Meanwhile - the best of luck.

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks for trying

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok steanson - I've got it first convert to sin and cos: sin3x/cos3x -six/cosx this can be written as (sin3xcosx - sinxcos3x) / cos3xcosx by the 'sums and differences of trig ratios formulae: the numerator = 1/2(sin4x +sin2x) - 1/2(sin4x- sin2x) which reduces to sin2x which we can write as 2sinxcosx sowe have: 2 sinxcosx/cos3xcosx cancelling out cosx = 2sinx/cos3x =2sinxsec3x

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the rules i referred to are sinx + siny = 2sin[(x+y/2][cos(x-y)/2] and sinx - siny = 2cos[(x+y/2][sin(x-y)/2] and I used them from' right to left' A good way to remember these is eg the first one 'sinx + siny = 2 sin semi sum cos semi difference' there are similar rules for the sum of differences of cosines. A higher grade maths book should have them

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks very much!, i think the reason i struggled is i didn't have those sinx + siny identities given in my text, handy to know!

  10. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.