A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Can someone PLEASE help me with 14d. and 15a. I have my precal final, and I still have no idea how to do some ?s!
anonymous
 3 years ago
Can someone PLEASE help me with 14d. and 15a. I have my precal final, and I still have no idea how to do some ?s!

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0does 15 d have an actual answer?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0tan^2(x)  sin^2(x) sin^2(x)/cos^2(x)  sin^2(x) [sin^2(x)  cos^2(x)sin^2(x)]/cos^2(x) sin^2(x)(1  cos^2(x))/cos^2(x) sin^2(x)sin^2(x)/cos^2(x) tan^2(x)sin^2(x)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but in this kind of problem both sides have to equal to one another, so in this casem @TweT226 is right. but why did you subtract it by 1?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but thank you for all the work @Hero your work was beautiful!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but the answer has to be equal to tan^2xsin^2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but that is the answer, tan^xsin^2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha yes youre right. lol thank you!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do u know how in his way why he subtracted by one though?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.014d: work from the right hand side

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.015a. \[\sin ( x  \Theta) = sinx \cos \theta  \sin \theta cosx \] then solve and put theta = pi \[\cos \Pi = 1 \] \[\sin \Pi = 0\] Put that into the equation and obtain sinx

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i just figured out 15a! thanks!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do u know how to change something in polar form? like sqrt(3) + i?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.014d. i will solve this from the right hand side of the equation: \[\tan ^2 x \sin^2x = \frac{ \sin^2x }{ \cos^2x } \times \left( \sin^2x \right)\] \[= \frac{ \sin^2x }{ \cos^2x } \times ( 1  \cos^2x)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0solve the equation and you will obtain the expression for the RHS

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i dont understand why you did (1−cos2x)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the identity of sin2x = 1cos2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so whenever you see that u change it to (1−cos2x)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0cause i thought that only applied when you added sin2x + cos2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well you can determine what you will need to convert depending on whats on the other side of the equation. right??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0solve and let me know how it goes. OK :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no it still didnt realy work out... here let me try again

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright, i will finish the solution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can i take a picture of what im doing so u can tell me what im doing wrong

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sin^2x\cos^xsin^2x }{ \cos^2x }\] \[\frac{ \sin^2x }{ \cos^2x }  \frac{ \cos^2xsin^2x }{ \cos^2x }\] \[\frac{ \sin^2x }{ \cos^2x }= \tan^2x\] and : \[\frac{ \cos^2xsin^2x }{ \cos^2x }= \sin^2x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sure you can! lemme check

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay cool! awh thanks!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I worked on from the RHS.. its much easier to figure out from the RHS but I will try to do it from the LHS as well. it should be the same answer. In the meantime, you can check the solutions i put up

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when u did sin2x−cosxsin2xcos2x why isnt it sin2x−cos2xsin2xcos2x?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why is it sin2x−cosxsin2xcos2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0from the top, we can see that its just a multiplication of sin2X and whats in the bracket

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah so shouldnt it be cos2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hang on, I'll upload the solution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know right.. :D got it from my computer.. mind reading it backwards?? :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0loll yeah ill try haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or u could take a picture on ur phone or something and email it to yo self! and did u find a common denominator?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0LaTex is horrible, i'm much of a "down on paper" breed//

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ugh ijust dont understand!!! im sorry :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh.. ok lets try that again:

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Solve from the Right Hand Side.. Did you get the solutions up to a certain point?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes, and each time i do it i stop at that point

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0at the sin2x  sin2xcos2x / cos2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0idk what to d after that because i get tan2x sin2xcos2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh there' the mistake right there.. you have to divide sin2x by Cos2x as well as divide by sin2xcos2x by cos2x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah but when i do it i dont have both of those cases thats all i have left

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok i will post the solution from that point

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when i solve that yeah i get the answer but i dont see how u got that :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol i agree with you but my teacher wont give me credit unless i do it that way

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0she told us if we do it any other way we wont get credit

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You are supposed to get credits for either solving from LHS or RHS

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah i agree im just confused

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i mean i honestly dont mind this way, just this problem..
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.