Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ianwalters

verify: secx - cosx = tanxsinx

  • one year ago
  • one year ago

  • This Question is Open
  1. whpalmer4
    Best Response
    You've already chosen the best response.
    Medals 2

    What's the definition of \(\sec x\)?

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

    cosx

    • one year ago
  3. whpalmer4
    Best Response
    You've already chosen the best response.
    Medals 2

    No. Why would we need a different name if it is the same?

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

    oh oops. its 1/cosx

    • one year ago
  5. whpalmer4
    Best Response
    You've already chosen the best response.
    Medals 2

    There we go. So what if you substitute 1/cos x in to the equation, what do you get?

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

    1/cosx - cosx = tanxsinx

    • one year ago
  7. whpalmer4
    Best Response
    You've already chosen the best response.
    Medals 2

    Okay, can you make a common denominator and combine 1/cos x - cos x?

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

    now: (1-cos^2x)/cosx

    • one year ago
  9. whpalmer4
    Best Response
    You've already chosen the best response.
    Medals 2

    okay, does that numerator look like any of your trig identities?

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

    yes but i cant remember what that is equivalent to

    • one year ago
  11. whpalmer4
    Best Response
    You've already chosen the best response.
    Medals 2

    well, the formula for a circle at the origin is \(x^2 + y^2 = r^2\) or if we are talking the unit circle used to define the trig functions, \x^2 + y^2 = 1\) and so \(\sin^2x + \cos^x = 1\) or \(\sin^2x = 1-\cos^2x\)

    • one year ago
  12. whpalmer4
    Best Response
    You've already chosen the best response.
    Medals 2

    rats, left out a '('! \(x^2+y^2=1\)

    • one year ago
  13. whpalmer4
    Best Response
    You've already chosen the best response.
    Medals 2

    So our left hand side is now \[\frac{\sin^2x}{\cos x}\]Wouldn't it be great if \[\tan x = \frac{\sin x}{\cos x}\] because that would make our right hand side be \[\frac{\sin x}{\cos x}*\sin x = \frac{\sin^2x}{\cos x}\]just like our left hand side. SOH CAH TOA sine = opposite over hypotenuse cosine = adjacent over hypotenuse tangent = opposite over adjacent Hey, look at that: opposite over hypotenuse divided by adjacent over hypotenuse = opposite over adjacent, so \[\tan x =\frac{\sin x}{\cos x}\]just like we want!

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

    thank you so much!

    • one year ago
    • Attachments:

See more questions >>>

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.