A community for students.
Here's the question you clicked on:
 0 viewing
ianwalters
 3 years ago
verify:
secx  cosx = tanxsinx
ianwalters
 3 years ago
verify: secx  cosx = tanxsinx

This Question is Open

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.2What's the definition of \(\sec x\)?

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.2No. Why would we need a different name if it is the same?

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.2There we go. So what if you substitute 1/cos x in to the equation, what do you get?

ianwalters
 3 years ago
Best ResponseYou've already chosen the best response.01/cosx  cosx = tanxsinx

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.2Okay, can you make a common denominator and combine 1/cos x  cos x?

ianwalters
 3 years ago
Best ResponseYou've already chosen the best response.0now: (1cos^2x)/cosx

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.2okay, does that numerator look like any of your trig identities?

ianwalters
 3 years ago
Best ResponseYou've already chosen the best response.0yes but i cant remember what that is equivalent to

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.2well, 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 \(\sin^2x = 1\cos^2x\)

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.2rats, left out a '('! \(x^2+y^2=1\)

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.2So 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!
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.