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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
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14. C 15.B
what?

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Other answers:

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

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