A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
PreCal Help Please..I am so stuck on what to do
anonymous
 3 years ago
PreCal Help Please..I am so stuck on what to do

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Use the equation below to find the value of sin x. \[\frac{ \tan x }{ \cot x }  \frac{ \sec x }{ \cos x }= \frac{ 2 }{ \csc x }\]

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0write everything in terms of sin and cos

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That is what I am not understanding. I have a really hard time writting out what it is supposed to be once you start changing everything.

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Hint: Write it this way. Then simply change everything to sin and cos \[\tan(x) \div \cot(x)  \sec(x) \div \cos(x) = 2 \div \csc(x)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So how would you change it?

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2\[\frac{\sin(x)}{\cos(x)} \div \frac{\cos(x)}{\sin(x)}  \frac{1}{\cos(x)} \div \cos(x) = 2 \div \frac{1}{\sin(x)}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so could the first two fracts cancel out?

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2How do you divide: \[\frac{1}{4} \div \frac{7}{6}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why not? BTW i missed the lessons over this stuff so i am really lost on almost all of it.

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Answer the previous question. How do you divide that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you switch the #'s of the second fraction so it is now 6/7 and you change divide to multiply. you now have 1/4 X 6/7 which is 6/28 simplified to 3/14

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Yes, so do that same thing with the trig equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it is now \[\frac{ \sin^2x }{ \cos^2x }\frac{ 1 }{ \cos x }\div \cos x=2\div \frac{ 1 }{ \sin x }\]??

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Yes, but do the rest of them. Change all of the divisions into multiplication.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it looks like\[\frac{ \sin^2x }{ \cos^2x }\frac{ 1 }{ cosx }\times \frac{ cosx }{ 1 }=2\times \frac{ sinx }{ 1 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If so, wouldn't 1/cosx times cosx/1 cancel out?

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2no, you did the cos(x)/1 incorrectly. cos(x) is already equal to cos(x)/1. You have to flip it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh duh ok so then it is \[\frac{ \sin^2 x}{ \cos^2x } \frac{ 1 }{ \cos x }\times \frac{ 1 }{ \cos x }= 2\times \frac{ sinx }{ 1 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so do the mult fractions combine to get 1/cos^2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so then what do you do?

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Yes, so what do you have after that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sin^2x }{ \cos^2x }\frac{ 1 }{ \cos^2x }= 2\times \frac{ sinx }{ 1 }\]

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Okay, do you know how to combine fractions on the left side?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0couldn't you turn 2x(sinx/1) into just 2sinx? and Im not sure what to do on the other side

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2On the left side, you have to combine fractions if they have the same denominator.

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2For example: \[\frac{1}{4} + \frac{x}{4} = \frac{1 + x}{4}\]

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2The general rule for combining fractions with the same denominator is: \[\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it is now \[\frac{ \sin^2x1 }{ \cos^2x }=2sinx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or is the 2sin x supposed to be sin^2x

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Now change the 1 to \(\sin^2x + \cos^2x\)

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2No, you have it right. 2sin(x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait, you put sin^2x + cos^2x instead of one? what is the new equation?

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Yes, you put \(\sin^2x + \cos^2x\) in place of 1 because they are equal.

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2What do you get afterwards?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sin^2x\sin^2x+\cos^2x }{ \cos^2x }=2sinx\] then you can subtract the sin^2x to get \[\frac{ \cos^2x }{ \cos^2x }=2sinx\] then if thats the case you would just have 2sinx

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Actually, if you do it correctly you get: \[\frac{ \cos^2x }{ \cos^2x }=2\sin x\]

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Which simplifies to \[1 = 2 \sin x\] Remember that we are still solving for x

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2You were supposed to get this after substituting the 1: \[\frac{ \sin^2x(\sin^2x+\cos^2x) }{ \cos^2x }=2\sin x\] Then get : \[\frac{ \sin^2x\sin^2xcos^2x }{ \cos^2x }=2\sin x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or no sinx would be 1/2

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Ultimately you end up isolating the sin(x) on the right side: \[\frac{1}{2} = \sin x\]

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Now all you have to do is take the inverse sin of both sides to get \[\sin^{1}\left(\frac{1}{2}\right) = x\]

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2So evaluate the left side using your calc. Let me know what you get. Also make sure it is in radian mode.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And thank you for your help :)

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Well, in exact mode, you should have gotten \[x = \frac{\pi}{6}\]

Hero
 3 years ago
Best ResponseYou've already chosen the best response.2Nevertheless:\[\frac{\pi}{6} \approx .52\]
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.