A community for students.
Here's the question you clicked on:
 0 viewing
Rosy95
 3 years ago
Please help on PreCal
Rosy95
 3 years ago
Please help on PreCal

This Question is Closed

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0Find an expression equivalent to... dw:1358280084977:dw

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1any expression? or is this something needing to involve a certain method? There's a number of ways to find an equivalent.

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0I am not sure. The answer choice are... A. sec^2 theta B. cot theta C. tan^2 theta D. cos^2 theta

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1358280261643:dw

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1358280353086:dw

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1Oh, my pictures got cut off, but the math is easy to see if you know definitions of trig functions and basic algebra skills.

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0Yes I can see. Thanks for writing out the equations. So the answer would be A. sec^2 theta?

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1358280516292:dw

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0So can you help with another one?

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1YEs. Post a new question

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0Simplifydw:1358281039137:dw

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1Tan^2theta + 1 = sec^2theta This is very useful to remember.

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0And can you check a question? The question is Which expression is equivalent to cos(pitheta)? I got cos theta. Is that right?

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1I'll answer that one quickly before I finish your first problem.

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1358281166638:dw

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1Another good rule to remember. So, yes your answer is correct.

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks and for my second question, I am still confused. Can you explain more? It was the simplify one

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1358281344118:dw

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1358281440642:dw

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0Oh ok. So since cos^2 theta can be crossed out, you move the 1st numerator over the 2nd denominator. That makes more sense. Thanks!

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1Well, it's not being crossed out. It's cos^2 theta divided by itself which is one. And then as fractions go, numerator times numerator and denominator times denominator.

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0Oh ok, Do you know how to do this one.... Solve \[4\sin ^{2}x+4\sqrt{2}cosx6=0\]

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1Post this one as a new question up top and close this question.
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.