Question

Prove that :
4cos2A*(cosA)^2=8(sinA)^2

Answer 1

i have been doing similar questions for the whole day, try using double-angle identities

(cosA)^2 =(cos^2 A) and that would be the same thing as (sinA)^2

Answer 2

I did that but unfortunately its useless ,i didn't reach the proof :(

Answer 3

This is an identity we might use:
\[\cos(2A)=\cos^2(A)-\sin^2(A)=(1-\sin^2(A))-\sin^2(A)=1-2\sin^2(A)\]
\[4(1-2\sin^2(A))\cos^2(A)=4(1-2\sin^2(A)(1-\sin^2(A))= \]
\[4(1-\sin^2(A)-2\sin^2(A)+\sin^4(A)) \]
\[4(1-3\sin^2(A)+\sin^4(A))\]
What you have isn't an identity

Answer 4

This is what I'm reading for what you have
\[4\cos(2A)(\cos(A))^2=8(\sin(A))^2\]
Is that what it said?

Answer 5

yes ,thats right @myininaya

Answer 6

We know this isn't an identity because if A=0
then
we have 4=0 which is not true

Answer 7

cos(0)=1
sin(0)=0

Answer 8

\[4\cos(2 \cdot 0) (\cos(0))^2=8(\sin(0))^2\]
\[4\cos(0)(\cos(0))^2=8(\sin(0))^2\]
\[4(1)(1)^2=8(0)^2\]
\[4=0 \]

Answer 9

not true so not an identity

Answer 10

but how do you know that A=0??

Answer 11

For this to be an identity it is suppose to be true for all values of A
There was a value of A that doesn't satisfy the equality so therefore it is not an identity
This is called a counterexample

Answer 12

OHH!! ok! so for future questions, i can use any number for A to prove and check whether the identity is true or not?

Answer 13

you cannot use it to prove
but you use it to disprove

Answer 14

so @myininaya ,whats da answer cuz i got confused .*_*

Answer 15

You don't understand that I plugged in 0 for A to show the identity is in fact not an identity

Answer 16

?

Answer 17

OHHH! ,ahaaa i got it

Answer 18

There are other values of A you could use to disprove also
I just chose A because it was the prettiest

Answer 19

But Eyad I still think what you have written is wrong

Answer 20

Since it says to prove

Answer 21

But I could be wrong about you being wrong

Answer 22

lol

Answer 23

lol,but in case thats its written right ,U have just said that we can't use that law (a=0) to prove ,just use it to disprove right ?

Answer 24

@myininaya : i got another question looks like this one ,maybe u can get a hint from it ,u want me to post it ?

Answer 25

no, @myininaya , i used the graphing calculator to check their graphs... it is not an identity.

Answer 26

@dpalnc Yes I know it is not an identity. I'm saying what he has made not be written correctly since it does say to prove it is an identity.