natasha.aries
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!
Delete
Share
This Question is Closed
natasha.aries
Best Response
You've already chosen the best response.
0
caesar27
Best Response
You've already chosen the best response.
0
14. C
15.B
natasha.aries
Best Response
You've already chosen the best response.
0
what?
natasha.aries
Best Response
You've already chosen the best response.
0
does 15 d have an actual answer?
natasha.aries
Best Response
You've already chosen the best response.
0
i mean a lol
TweT226
Best Response
You've already chosen the best response.
0
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
Best Response
You've already chosen the best response.
0
\[\tan^2x - \sin^2x = \tan^2x\sin^2x\]
\[\frac{\tan^2x - \sin^2x}{\sin^2x} = \tan^2x\]
\[\frac{\tan^2x}{\sin^2x} - \frac{\sin^2x}{\sin^2x} = \tan^2x\]
\[\frac{\tan^2x}{\sin^2x} - 1 = \tan^2x\]
\[\frac{\tan^2x}{\sin^2x} = \tan^2x + 1\]
\[\frac{\tan^2x}{\sin^2x} = \frac{\sin^2x}{\cos^2x} + \frac{\cos^2x}{\cos^2x}\]
\[\frac{\tan^2x}{\sin^2x} = \frac{\sin^2x+\cos^2x}{\cos^2x}\]
\[\frac{\tan^2x}{\sin^2x} = \frac{1}{\cos^2x}\]
\[\tan^2x = \frac{\sin^2x}{\cos^2x}\]
\[\tan^2x = \tan^2x\]
natasha.aries
Best Response
You've already chosen the best response.
0
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?
natasha.aries
Best Response
You've already chosen the best response.
0
but thank you for all the work @Hero your work was beautiful!
natasha.aries
Best Response
You've already chosen the best response.
0
but the answer has to be equal to tan^2xsin^2x
natasha.aries
Best Response
You've already chosen the best response.
0
but that is the answer, tan^xsin^2x
natasha.aries
Best Response
You've already chosen the best response.
0
haha yes youre right. lol thank you!
natasha.aries
Best Response
You've already chosen the best response.
0
do u know how in his way why he subtracted by one though?
TweT226
Best Response
You've already chosen the best response.
0
mine or hero?
natasha.aries
Best Response
You've already chosen the best response.
0
yours
cesckay
Best Response
You've already chosen the best response.
0
14d: work from the right hand side
cesckay
Best Response
You've already chosen the best response.
0
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
natasha.aries
Best Response
You've already chosen the best response.
0
i just figured out 15a! thanks!
natasha.aries
Best Response
You've already chosen the best response.
0
do u know how to change something in polar form? like sqrt(3) + i?
cesckay
Best Response
You've already chosen the best response.
0
14d.
i will solve this from the right hand side of the equation:
\[\tan ^2 x \sin^2x = \frac{ \sin^2x }{ \cos^2x } \times \left( \sin^2x \right)\]
\[= \frac{ \sin^2x }{ \cos^2x } \times ( 1 - \cos^2x)\]
cesckay
Best Response
You've already chosen the best response.
0
solve the equation and you will obtain the expression for the RHS
natasha.aries
Best Response
You've already chosen the best response.
0
i dont understand why you did (1−cos2x)
cesckay
Best Response
You've already chosen the best response.
0
the identity of sin2x = 1-cos2x
natasha.aries
Best Response
You've already chosen the best response.
0
so whenever you see that u change it to (1−cos2x)?
natasha.aries
Best Response
You've already chosen the best response.
0
cause i thought that only applied when you added sin2x + cos2x
cesckay
Best Response
You've already chosen the best response.
0
well you can determine what you will need to convert depending on whats on the other side of the equation. right??
cesckay
Best Response
You've already chosen the best response.
0
solve and let me know how it goes. OK :)
natasha.aries
Best Response
You've already chosen the best response.
0
okay! hold on!
natasha.aries
Best Response
You've already chosen the best response.
0
no it still didnt realy work out...
here let me try again
cesckay
Best Response
You've already chosen the best response.
0
alright, i will finish the solution
natasha.aries
Best Response
You've already chosen the best response.
0
can i take a picture of what im doing so u can tell me what im doing wrong
cesckay
Best Response
You'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 :
\[-\frac{ \cos^2xsin^2x }{ \cos^2x }= -\sin^2x\]
cesckay
Best Response
You've already chosen the best response.
0
sure you can!
lemme check
natasha.aries
Best Response
You've already chosen the best response.
0
okay cool! awh thanks!
natasha.aries
Best Response
You've already chosen the best response.
0
hold onn!
natasha.aries
Best Response
You've already chosen the best response.
0
cesckay
Best Response
You've already chosen the best response.
0
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
natasha.aries
Best Response
You've already chosen the best response.
0
okay thank youuu
natasha.aries
Best Response
You've already chosen the best response.
0
when u did sin2x−cosxsin2xcos2x
why isnt it sin2x−cos2xsin2xcos2x?
natasha.aries
Best Response
You've already chosen the best response.
0
why is it sin2x−cosxsin2xcos2x
cesckay
Best Response
You've already chosen the best response.
0
from the top, we can see that its just a multiplication of sin2X and whats in the bracket
natasha.aries
Best Response
You've already chosen the best response.
0
yeah so shouldnt it be cos2x
cesckay
Best Response
You've already chosen the best response.
0
hang on, I'll upload the solution
cesckay
Best Response
You've already chosen the best response.
0
cesckay
Best Response
You've already chosen the best response.
0
i know right.. :D got it from my computer.. mind reading it backwards?? :/
natasha.aries
Best Response
You've already chosen the best response.
0
loll yeah ill try haha
natasha.aries
Best Response
You've already chosen the best response.
0
or u could take a picture on ur phone or something and email it to yo self!
and did u find a common denominator?
cesckay
Best Response
You've already chosen the best response.
0
ook yeah try this
cesckay
Best Response
You've already chosen the best response.
0
LaTex is horrible, i'm much of a "down on paper" breed//
natasha.aries
Best Response
You've already chosen the best response.
0
ugh ijust dont understand!!! im sorry :(
cesckay
Best Response
You've already chosen the best response.
0
Oh.. ok lets try that again:
cesckay
Best Response
You've already chosen the best response.
0
Solve from the Right Hand Side.. Did you get the solutions up to a certain point?
natasha.aries
Best Response
You've already chosen the best response.
0
yes, and each time i do it i stop at that point
cesckay
Best Response
You've already chosen the best response.
0
where exactly?
natasha.aries
Best Response
You've already chosen the best response.
0
at the sin2x - sin2xcos2x / cos2x
natasha.aries
Best Response
You've already chosen the best response.
0
idk what to d after that because i get tan2x- sin2xcos2x
cesckay
Best Response
You've already chosen the best response.
0
Oh there' the mistake right there..
you have to divide sin2x by Cos2x
as well as divide by -sin2xcos2x by cos2x
natasha.aries
Best Response
You've already chosen the best response.
0
yeah but when i do it i dont have both of those cases thats all i have left
cesckay
Best Response
You've already chosen the best response.
0
Ok i will post the solution from that point
cesckay
Best Response
You've already chosen the best response.
0
\[\frac{ \sin^2x-\sin^2xcos^2x }{ \cos^2x }\]
cesckay
Best Response
You've already chosen the best response.
0
\[\frac{ \sin^2x }{ \cos^2x }-\frac{ \sin^2xcos^2x }{ \cos^2x }\]
cesckay
Best Response
You've already chosen the best response.
0
solve that..
natasha.aries
Best Response
You've already chosen the best response.
0
thanks @Hero !!!
natasha.aries
Best Response
You've already chosen the best response.
0
when i solve that yeah i get the answer but i dont see how u got that :(
natasha.aries
Best Response
You've already chosen the best response.
0
lol i agree with you but my teacher wont give me credit unless i do it that way
natasha.aries
Best Response
You've already chosen the best response.
0
she told us if we do it any other way we wont get credit
cesckay
Best Response
You've already chosen the best response.
0
You are supposed to get credits for either solving from LHS or RHS
natasha.aries
Best Response
You've already chosen the best response.
0
yeah i agree im just confused
natasha.aries
Best Response
You've already chosen the best response.
0
i mean i honestly dont mind this way, just this problem..