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anonymous
 4 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
 4 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!

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

anonymous
 4 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
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0but thank you for all the work @Hero your work was beautiful!

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

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

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

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

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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0i just figured out 15a! thanks!

anonymous
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0solve the equation and you will obtain the expression for the RHS

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

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

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

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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0solve and let me know how it goes. OK :)

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

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

anonymous
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0sure you can! lemme check

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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0when u did sin2x−cosxsin2xcos2x why isnt it sin2x−cos2xsin2xcos2x?

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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0yeah so shouldnt it be cos2x

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

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

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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0LaTex is horrible, i'm much of a "down on paper" breed//

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

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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0yes, and each time i do it i stop at that point

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

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

anonymous
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0Ok i will post the solution from that point

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

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

anonymous
 4 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
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.0You are supposed to get credits for either solving from LHS or RHS

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

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