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natasha.aries
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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!
 2 years ago
 2 years ago
natasha.aries Group Title
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!
 2 years ago
 2 years ago

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caesar27 Group TitleBest ResponseYou've already chosen the best response.0
14. C 15.B
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
what?
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
does 15 d have an actual answer?
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
i mean a lol
 2 years ago

TweT226 Group TitleBest ResponseYou'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)
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
\[\tan^2x  \sin^2x = \tan^2x\sin^2x\] \[\frac{\tan^2x  \sin^2x}{\sin^2x} = \tan^2x\] \[\frac{\tan^2x}{\sin^2x}  \frac{\sin^2x}{\sin^2x} = \tan^2x\] \[\frac{\tan^2x}{\sin^2x}  1 = \tan^2x\] \[\frac{\tan^2x}{\sin^2x} = \tan^2x + 1\] \[\frac{\tan^2x}{\sin^2x} = \frac{\sin^2x}{\cos^2x} + \frac{\cos^2x}{\cos^2x}\] \[\frac{\tan^2x}{\sin^2x} = \frac{\sin^2x+\cos^2x}{\cos^2x}\] \[\frac{\tan^2x}{\sin^2x} = \frac{1}{\cos^2x}\] \[\tan^2x = \frac{\sin^2x}{\cos^2x}\] \[\tan^2x = \tan^2x\]
 2 years ago

natasha.aries Group TitleBest ResponseYou'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?
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
but thank you for all the work @Hero your work was beautiful!
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
but the answer has to be equal to tan^2xsin^2x
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
but that is the answer, tan^xsin^2x
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
haha yes youre right. lol thank you!
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
do u know how in his way why he subtracted by one though?
 2 years ago

TweT226 Group TitleBest ResponseYou've already chosen the best response.0
mine or hero?
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
yours
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
14d: work from the right hand side
 2 years ago

cesckay Group TitleBest ResponseYou'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
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
i just figured out 15a! thanks!
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
do u know how to change something in polar form? like sqrt(3) + i?
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
14d. i will solve this from the right hand side of the equation: \[\tan ^2 x \sin^2x = \frac{ \sin^2x }{ \cos^2x } \times \left( \sin^2x \right)\] \[= \frac{ \sin^2x }{ \cos^2x } \times ( 1  \cos^2x)\]
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
solve the equation and you will obtain the expression for the RHS
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
i dont understand why you did (1−cos2x)
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
the identity of sin2x = 1cos2x
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
so whenever you see that u change it to (1−cos2x)?
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
cause i thought that only applied when you added sin2x + cos2x
 2 years ago

cesckay Group TitleBest ResponseYou'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??
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
solve and let me know how it goes. OK :)
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
okay! hold on!
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
no it still didnt realy work out... here let me try again
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
alright, i will finish the solution
 2 years ago

natasha.aries Group TitleBest ResponseYou'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
 2 years ago

cesckay Group TitleBest 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 : \[\frac{ \cos^2xsin^2x }{ \cos^2x }= \sin^2x\]
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
sure you can! lemme check
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
okay cool! awh thanks!
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
hold onn!
 2 years ago

cesckay Group TitleBest ResponseYou'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
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
okay thank youuu
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
when u did sin2x−cosxsin2xcos2x why isnt it sin2x−cos2xsin2xcos2x?
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
why is it sin2x−cosxsin2xcos2x
 2 years ago

cesckay Group TitleBest ResponseYou'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
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
yeah so shouldnt it be cos2x
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
hang on, I'll upload the solution
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
i know right.. :D got it from my computer.. mind reading it backwards?? :/
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
loll yeah ill try haha
 2 years ago

natasha.aries Group TitleBest ResponseYou'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?
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
ook yeah try this
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
LaTex is horrible, i'm much of a "down on paper" breed//
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
ugh ijust dont understand!!! im sorry :(
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
Oh.. ok lets try that again:
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
Solve from the Right Hand Side.. Did you get the solutions up to a certain point?
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
yes, and each time i do it i stop at that point
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
where exactly?
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
at the sin2x  sin2xcos2x / cos2x
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
idk what to d after that because i get tan2x sin2xcos2x
 2 years ago

cesckay Group TitleBest ResponseYou'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
 2 years ago

natasha.aries Group TitleBest ResponseYou'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
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
Ok i will post the solution from that point
 2 years ago

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

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

cesckay Group TitleBest ResponseYou've already chosen the best response.0
solve that..
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
thanks @Hero !!!
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
when i solve that yeah i get the answer but i dont see how u got that :(
 2 years ago

natasha.aries Group TitleBest ResponseYou'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
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
she told us if we do it any other way we wont get credit
 2 years ago

cesckay Group TitleBest ResponseYou've already chosen the best response.0
You are supposed to get credits for either solving from LHS or RHS
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
yeah i agree im just confused
 2 years ago

natasha.aries Group TitleBest ResponseYou've already chosen the best response.0
i mean i honestly dont mind this way, just this problem..
 2 years ago
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