anonymous
  • anonymous
Can someone PLEASE help me with 14d. and 15a. I have my pre-cal final, and I still have no idea how to do some ?s!
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
14. C 15.B
anonymous
  • anonymous
what?

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anonymous
  • anonymous
does 15 d have an actual answer?
anonymous
  • anonymous
i mean a lol
anonymous
  • anonymous
tan^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
  • Hero
\[\tan^2x - \sin^2x = \tan^2x\sin^2x\] \[\frac{\tan^2x - \sin^2x}{\sin^2x} = \tan^2x\] \[\frac{\tan^2x}{\sin^2x} - \frac{\sin^2x}{\sin^2x} = \tan^2x\] \[\frac{\tan^2x}{\sin^2x} - 1 = \tan^2x\] \[\frac{\tan^2x}{\sin^2x} = \tan^2x + 1\] \[\frac{\tan^2x}{\sin^2x} = \frac{\sin^2x}{\cos^2x} + \frac{\cos^2x}{\cos^2x}\] \[\frac{\tan^2x}{\sin^2x} = \frac{\sin^2x+\cos^2x}{\cos^2x}\] \[\frac{\tan^2x}{\sin^2x} = \frac{1}{\cos^2x}\] \[\tan^2x = \frac{\sin^2x}{\cos^2x}\] \[\tan^2x = \tan^2x\]
anonymous
  • anonymous
but 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
  • anonymous
but thank you for all the work @Hero your work was beautiful!
anonymous
  • anonymous
but the answer has to be equal to tan^2xsin^2x
anonymous
  • anonymous
but that is the answer, tan^xsin^2x
anonymous
  • anonymous
haha yes youre right. lol thank you!
anonymous
  • anonymous
do u know how in his way why he subtracted by one though?
anonymous
  • anonymous
mine or hero?
anonymous
  • anonymous
yours
anonymous
  • anonymous
14d: work from the right hand side
anonymous
  • anonymous
15a. \[\sin ( x - \Theta) = sinx \cos \theta - \sin \theta cosx \] then solve and put theta = pi \[\cos \Pi = -1 \] \[\sin \Pi = 0\] Put that into the equation and obtain -sinx
anonymous
  • anonymous
i just figured out 15a! thanks!
anonymous
  • anonymous
do u know how to change something in polar form? like sqrt(3) + i?
anonymous
  • anonymous
14d. 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
  • anonymous
solve the equation and you will obtain the expression for the RHS
anonymous
  • anonymous
i dont understand why you did (1−cos2x)
anonymous
  • anonymous
the identity of sin2x = 1-cos2x
anonymous
  • anonymous
so whenever you see that u change it to (1−cos2x)?
anonymous
  • anonymous
cause i thought that only applied when you added sin2x + cos2x
anonymous
  • anonymous
well you can determine what you will need to convert depending on whats on the other side of the equation. right??
anonymous
  • anonymous
solve and let me know how it goes. OK :)
anonymous
  • anonymous
okay! hold on!
anonymous
  • anonymous
no it still didnt realy work out... here let me try again
anonymous
  • anonymous
alright, i will finish the solution
anonymous
  • anonymous
can i take a picture of what im doing so u can tell me what im doing wrong
anonymous
  • anonymous
\[\frac{ \sin^2x-\cos^xsin^2x }{ \cos^2x }\] \[\frac{ \sin^2x }{ \cos^2x } - \frac{ \cos^2xsin^2x }{ \cos^2x }\] \[\frac{ \sin^2x }{ \cos^2x }= \tan^2x\] and : \[-\frac{ \cos^2xsin^2x }{ \cos^2x }= -\sin^2x\]
anonymous
  • anonymous
sure you can! lemme check
anonymous
  • anonymous
okay cool! awh thanks!
anonymous
  • anonymous
hold onn!
anonymous
  • anonymous
anonymous
  • anonymous
I 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
  • anonymous
okay thank youuu
anonymous
  • anonymous
when u did sin2x−cosxsin2xcos2x why isnt it sin2x−cos2xsin2xcos2x?
anonymous
  • anonymous
why is it sin2x−cosxsin2xcos2x
anonymous
  • anonymous
from the top, we can see that its just a multiplication of sin2X and whats in the bracket
anonymous
  • anonymous
yeah so shouldnt it be cos2x
anonymous
  • anonymous
hang on, I'll upload the solution
anonymous
  • anonymous
anonymous
  • anonymous
i know right.. :D got it from my computer.. mind reading it backwards?? :/
anonymous
  • anonymous
loll yeah ill try haha
anonymous
  • anonymous
or u could take a picture on ur phone or something and email it to yo self! and did u find a common denominator?
anonymous
  • anonymous
ook yeah try this
anonymous
  • anonymous
LaTex is horrible, i'm much of a "down on paper" breed//
anonymous
  • anonymous
ugh ijust dont understand!!! im sorry :(
anonymous
  • anonymous
Oh.. ok lets try that again:
anonymous
  • anonymous
Solve from the Right Hand Side.. Did you get the solutions up to a certain point?
anonymous
  • anonymous
yes, and each time i do it i stop at that point
anonymous
  • anonymous
where exactly?
anonymous
  • anonymous
at the sin2x - sin2xcos2x / cos2x
anonymous
  • anonymous
idk what to d after that because i get tan2x- sin2xcos2x
anonymous
  • anonymous
Oh there' the mistake right there.. you have to divide sin2x by Cos2x as well as divide by -sin2xcos2x by cos2x
anonymous
  • anonymous
yeah but when i do it i dont have both of those cases thats all i have left
anonymous
  • anonymous
Ok i will post the solution from that point
anonymous
  • anonymous
\[\frac{ \sin^2x-\sin^2xcos^2x }{ \cos^2x }\]
anonymous
  • anonymous
\[\frac{ \sin^2x }{ \cos^2x }-\frac{ \sin^2xcos^2x }{ \cos^2x }\]
anonymous
  • anonymous
solve that..
anonymous
  • anonymous
thanks @Hero !!!
anonymous
  • anonymous
when i solve that yeah i get the answer but i dont see how u got that :(
anonymous
  • anonymous
lol i agree with you but my teacher wont give me credit unless i do it that way
anonymous
  • anonymous
she told us if we do it any other way we wont get credit
anonymous
  • anonymous
You are supposed to get credits for either solving from LHS or RHS
anonymous
  • anonymous
yeah i agree im just confused
anonymous
  • anonymous
i mean i honestly dont mind this way, just this problem..

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