Question

True or False?

tan^2x= 1+ cos2x / 1-cos2x

Answer 1

\(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\)

Answer 2

verify or prove ?

Answer 3

Multiply both sides by 1-cos^2 x and simplify , keeping in mind that \[tan(x)=\frac{sin(x)}{cos(x)}\]

Answer 4

It does not specify on the question itself but I would assume proving.

Answer 5

(verify = manipulate 1 side at a time only
prove= both sides can be manipulate)

this is a traditional interpretation/definition of these words.

Answer 6

ok, lets just in case verify it.
will play 1 side at a time

Answer 7

Ok

Answer 8

\(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\)

can you re-write the bottom ?

Answer 9

use the fact that \(\large\color{black}{ \displaystyle \sin^2(x)+\cos^2(x)=1 }\)

Answer 10

So you would be able to re-write the bottem

Answer 11

yes, in the fact I just offered, subtract cos^2(x) from both sides, and tell me what you get

Answer 12

From the sin^2(x) one?

Answer 13

yes, from the sin^2(x)+cos^2(x)=1, there subtract cos^2(x) from both sides

Answer 14

\[\sin ^{2}(x)=-\cos ^{2}(x)+1\]

Answer 15

yes sin^2(x)=1-cos^2(x)

Answer 16

now, we have

\(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\)

can you re-write the bottom for me?

Answer 17

That is false
As 1+cos2x=1+2cos^2x-1
=2cos^2x
And 1-cos2x=1-(1-2sin^2x)
=2sin^2x
Put these values you will get the answer.

Answer 18

just put x = pi/4 , you can see how it is false

Answer 19

When I rearrange this @SolomonZelman, will I be taking the -cos^2(x) away from the denominator and onto the numerator?

I am not very good at trig

Answer 20

\(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\)

\(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{\sin^2(x)} }\)

that is what I meant

Answer 21

anyway, i think plugging in value to disprove is valid enough.

Answer 22

unless u want to go and do it without plug ins

Answer 23

Would I just have to plug in pi/4 to the equation?

Answer 24

But that's not a proper method

Answer 25

Oh ok

Answer 26

Well thank you guys for the help and time

Answer 27

Who said : proving the invalid of an expression by a counterexample is not a proper method?

Answer 28

That is the MOST proper method to prove something wrong.

Answer 29

\(\large\color{black}{ \displaystyle \tan^2(x)=\frac{ 1+\cos^2(x) }{1-\cos^2(x)} }\)


\(\large\color{black}{ \displaystyle \frac{\sin^2x}{\cos^2x}=\frac{ (1+\cos^2x) }{(1-\cos^2x)} }\)


\(\large\color{black}{ \displaystyle \frac{\sin^2x}{\cos^2x}=\frac{ (1+\cos^2x) }{(1-\cos^2x)} }\)


\(\large\color{black}{ \displaystyle \sin^2x(1-\cos^2x)=\cos^2x(1+\cos^2x) }\)


\(\large\color{black}{ \displaystyle \sin^2x-\sin^2x\cos^2x=\cos^2x+\cos^2x\cos^2x}\)


\(\large\color{black}{ \displaystyle \sin^2x=\sin^2x\cos^2x+\cos^2x+\cos^2x\cos^2x}\)


\(\large\color{black}{ \displaystyle \sin^2x=\cos^2x(\sin^2x+1+\cos^2x)}\)


\(\large\color{black}{ \displaystyle \sin^2x=\cos^2x(2)}\)


\(\large\color{black}{ \displaystyle \sin^2x=2(1- \sin^2x)}\)

\(\large\color{black}{ \displaystyle \sin^2x=2- 2\sin^2x}\)

\(\large\color{black}{ \displaystyle 3\sin^2x=2}\)

\(\large\color{black}{ \displaystyle \sin^2x=2/3}\)

\(\large\color{black}{ \displaystyle \sin x=\pm\sqrt{2/3}}\)

\(\large\color{black}{ \displaystyle x=\sin^{-1}\left(\pm\sqrt{2/3}\right)}\)

Answer 30

if your definition were to be true for all numbers, then this solution for x would be such that includes all numbers.

\(\large\color{black}{ \displaystyle \sin^{-1}\left(\pm\sqrt{2/3}\right)}\) is no way a representation of all [real] numbers!

Answer 31

*Takes a picture for future reference*

Thank you for the explanation! This will help me 200x better than before!

Answer 32

i took everything down piece by piece. it isn't that bad usually though