Question

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!

Answer 1

14. C
15.B

Answer 2

what?

Answer 3

does 15 d have an actual answer?

Answer 4

i mean a lol

Answer 5

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)

Answer 6

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

Answer 7

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?

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

haha yes youre right. lol thank you!

Answer 12

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

Answer 13

mine or hero?

Answer 14

yours

Answer 15

14d: work from the right hand side

Answer 16

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

Answer 17

i just figured out 15a! thanks!

Answer 18

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

Answer 19

14d.
i will solve this from the right hand side of the equation:
\[\tan ^2 x \sin^2x = \frac{ \sin^2x }{ \cos^2x } \times \left( \sin^2x \right)\]
\[= \frac{ \sin^2x }{ \cos^2x } \times ( 1 - \cos^2x)\]

Answer 20

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

Answer 21

i dont understand why you did (1−cos2x)

Answer 22

the identity of sin2x = 1-cos2x

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

okay! hold on!

Answer 28

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

Answer 29

alright, i will finish the solution

Answer 30

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

Answer 31

\[\frac{ \sin^2x-\cos^xsin^2x }{ \cos^2x }\]
\[\frac{ \sin^2x }{ \cos^2x } - \frac{ \cos^2xsin^2x }{ \cos^2x }\]
\[\frac{ \sin^2x }{ \cos^2x }= \tan^2x\]
and :
\[-\frac{ \cos^2xsin^2x }{ \cos^2x }= -\sin^2x\]

Answer 32

sure you can!
lemme check

Answer 33

okay cool! awh thanks!

Answer 34

hold onn!

Answer 35

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

Answer 36

okay thank youuu

Answer 37

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

Answer 38

why is it sin2x−cosxsin2xcos2x

Answer 39

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

Answer 40

yeah so shouldnt it be cos2x

Answer 41

hang on, I'll upload the solution

Answer 42

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

Answer 43

loll yeah ill try haha

Answer 44

or u could take a picture on ur phone or something and email it to yo self!
and did u find a common denominator?

Answer 45

ook yeah try this

Answer 46

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

Answer 47

ugh ijust dont understand!!! im sorry :(

Answer 48

Oh.. ok lets try that again:

Answer 49

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

Answer 50

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

Answer 51

where exactly?

Answer 52

at the sin2x - sin2xcos2x / cos2x

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

Ok i will post the solution from that point

Answer 57

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

Answer 58

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

Answer 59

solve that..

Answer 60

thanks @Hero !!!

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

yeah i agree im just confused

Answer 66

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