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!

- anonymous

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- anonymous

##### 1 Attachment

- anonymous

14. C
15.B

- anonymous

what?

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## More answers

- anonymous

does 15 d have an actual answer?

- anonymous

i mean a lol

- 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

\[\tan^2x - \sin^2x = \tan^2x\sin^2x\]
\[\frac{\tan^2x - \sin^2x}{\sin^2x} = \tan^2x\]
\[\frac{\tan^2x}{\sin^2x} - \frac{\sin^2x}{\sin^2x} = \tan^2x\]
\[\frac{\tan^2x}{\sin^2x} - 1 = \tan^2x\]
\[\frac{\tan^2x}{\sin^2x} = \tan^2x + 1\]
\[\frac{\tan^2x}{\sin^2x} = \frac{\sin^2x}{\cos^2x} + \frac{\cos^2x}{\cos^2x}\]
\[\frac{\tan^2x}{\sin^2x} = \frac{\sin^2x+\cos^2x}{\cos^2x}\]
\[\frac{\tan^2x}{\sin^2x} = \frac{1}{\cos^2x}\]
\[\tan^2x = \frac{\sin^2x}{\cos^2x}\]
\[\tan^2x = \tan^2x\]

- anonymous

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

but thank you for all the work @Hero your work was beautiful!

- anonymous

but the answer has to be equal to tan^2xsin^2x

- anonymous

but that is the answer, tan^xsin^2x

- anonymous

haha yes youre right. lol thank you!

- anonymous

do u know how in his way why he subtracted by one though?

- anonymous

mine or hero?

- anonymous

yours

- anonymous

14d: work from the right hand side

- 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

i just figured out 15a! thanks!

- anonymous

do u know how to change something in polar form? like sqrt(3) + i?

- 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

solve the equation and you will obtain the expression for the RHS

- anonymous

i dont understand why you did (1−cos2x)

- anonymous

the identity of sin2x = 1-cos2x

- anonymous

so whenever you see that u change it to (1−cos2x)?

- anonymous

cause i thought that only applied when you added sin2x + cos2x

- anonymous

well you can determine what you will need to convert depending on whats on the other side of the equation. right??

- anonymous

solve and let me know how it goes. OK :)

- anonymous

okay! hold on!

- anonymous

no it still didnt realy work out...
here let me try again

- anonymous

alright, i will finish the solution

- anonymous

can i take a picture of what im doing so u can tell me what im doing wrong

- 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

sure you can!
lemme check

- anonymous

okay cool! awh thanks!

- anonymous

hold onn!

- anonymous

##### 1 Attachment

- 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

okay thank youuu

- anonymous

when u did sin2x−cosxsin2xcos2x
why isnt it sin2x−cos2xsin2xcos2x?

- anonymous

why is it sin2x−cosxsin2xcos2x

- anonymous

from the top, we can see that its just a multiplication of sin2X and whats in the bracket

- anonymous

yeah so shouldnt it be cos2x

- anonymous

hang on, I'll upload the solution

- anonymous

##### 2 Attachments

- anonymous

i know right.. :D got it from my computer.. mind reading it backwards?? :/

- anonymous

loll yeah ill try haha

- 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

ook yeah try this

##### 2 Attachments

- anonymous

LaTex is horrible, i'm much of a "down on paper" breed//

- anonymous

ugh ijust dont understand!!! im sorry :(

- anonymous

Oh.. ok lets try that again:

- anonymous

Solve from the Right Hand Side.. Did you get the solutions up to a certain point?

- anonymous

yes, and each time i do it i stop at that point

- anonymous

where exactly?

- anonymous

at the sin2x - sin2xcos2x / cos2x

- anonymous

idk what to d after that because i get tan2x- sin2xcos2x

- anonymous

Oh there' the mistake right there..
you have to divide sin2x by Cos2x
as well as divide by -sin2xcos2x by cos2x

- anonymous

yeah but when i do it i dont have both of those cases thats all i have left

- anonymous

Ok i will post the solution from that point

- anonymous

\[\frac{ \sin^2x-\sin^2xcos^2x }{ \cos^2x }\]

- anonymous

\[\frac{ \sin^2x }{ \cos^2x }-\frac{ \sin^2xcos^2x }{ \cos^2x }\]

- anonymous

solve that..

- anonymous

thanks @Hero !!!

- anonymous

when i solve that yeah i get the answer but i dont see how u got that :(

- anonymous

lol i agree with you but my teacher wont give me credit unless i do it that way

- anonymous

she told us if we do it any other way we wont get credit

- anonymous

You are supposed to get credits for either solving from LHS or RHS

- anonymous

yeah i agree im just confused

- anonymous

i mean i honestly dont mind this way, just this problem..

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