A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ParthKohli

  • one year ago

Hey, this problem has a really elegant solution.

  • This Question is Closed
  1. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If \(m^2 + n^2 = 1\), \(p^2 + q^2 = 1\) and \(mp + nq = 0\) then what is \(mn + pq\)?

  2. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Meh - there's actually nothing elegant about it. The guy just made it look like that. But still, it looks really clever and let's see if you guys can come up with it.

  3. AbdullahM
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I assume we will need to use trignometry to solve it?

  4. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That sounds like a good way to do it... in the system of real numbers.

  5. rational
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    nvm was working assuming mp+nq=1

  6. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Haha, I'm really sorry about that.

  7. AbdullahM
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\sf m^2+n^2=1\) There is a trig identity: \(\sf cos^2(x)+sin^2(x)=1\) So then: \(\sf m= cos~(\alpha)\) \(\sf n=sin~(\alpha)\) Do the same for: \(\sf p^2+q^2=1\) \(\sf p= cos~(\beta)\) \(\sf q=sin~(\beta)\) Now, they gave us mp+nq=0 Substitute the values we got above into this second equation: \(\sf mp+nq=0\) \(\sf cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)=0\) And there is a trig identity: \(\sf cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)=cos(\alpha-\beta)\) so we can simplify that further down to: \(\sf cos(\alpha-\beta)=0\) I hope I'm going in the correct direction so far xD

  8. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, of course. But don't forget that there's a huge world of things that aren't real numbers.

  9. rational
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Nice! here is another alternative that looks kinda neat \[mn+pq=mn(p^2+q^2)+(m^2+n^2)pq =(mp+nq)(np+mq)=0\]

  10. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    That's exactly the solution I was about to post. Great job.

  11. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Where did you find this?

  12. rational
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    copied it from quora, there are plenty of solutions there but i liked that one the most

  13. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Haha, Anders Kaseorg. Same.

  14. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    He's just brilliant. I've been stalking him for a few days.

  15. AbdullahM
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oooh, interesting :o

  16. mathmath333
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    looks somewhat similar to this https://en.wikipedia.org/wiki/Brahmagupta's_identity

  17. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Haha, yes. :)

  18. 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.