Question

why is this true?

\[\cos(2\times2t)=2\cos ^{2}2t-1\]

Answer 1

cos(a+b)= cos(a)cos(b) - sin(a) sin(b)

if a and b are the same thing i.e. b= a then this is
cos(a+a)= cos^2(a) - sin^2(a)

also, you might know that
sin^2(a) + cos^(a) =1
you can use this to replace the sin^2(a) in the above equation

Answer 2

where did this come from?
\[\cos (a+b)\]

Answer 3

You might think of cos(a+b) =cos(a+a) when b=a. So, cos(a+a)=cos(2a) and you can apply the formula for cos(a+b)=cosacosb-sinasinb and then replace all your b's with a's. Once you have that, replace your a with 2t and you have eyour answer.

Answer 4

the cos of the sum of 2 angles cos(a+b) is well-known
and can be proven many ways.

see http://www.proofwiki.org/wiki/Cosine_of_Sum

Answer 5

ok, that is not what my confusing is, my original LHS was cos (2 times 2t), how did that turned into cos (2+2t)?

Answer 6

2* x is the same as x+x

in your case 2*2t is the same as 2t + 2t

if we call 2t = a
we can use cos(a+a)

Answer 7

ok i got you, this is good new information, thanks!

Answer 8

can you finish ?

Answer 9

i take a look and let you know in a minute...

Answer 10

in case the typo confuses you
also, you might know that
\[ \sin^2(a) + \cos^2(a) =1 \]
you can use this to replace the sin^2(a) in the above equation

Answer 11

ok i got it, i see, now i just have to work out the rest of the problem which is a verification problem, this was just one of its step, i might need you help, let you know, thanks so far!

Answer 12

ok, im completely lost in all the formulas

Answer 13

original question was
\[\cos 4t = 8\cos ^{4}t-8\cos ^{2}t+1\]

Answer 14

i got as far as you explained

Answer 15

now im stuck at
\[2 \cos ^{2}2t-1\]

Answer 16

That's your left hand side. Your right hand side is in terms of powers of cosines, so see if you can get them all together and simplify it, perhaps look at factoring. That is usually the case when we are trying to solve a trigonometric equation.

Answer 17

i need to verify that both side equal, according to the answer the RHS doesn't change

Answer 18

Oh, forgot you are trying to verify an identity.

Answer 19

In that case, I would have started with the more complicated side, which is the RHS.

Answer 20

The right hand side factors, btw.

Answer 21

ok im listening to your idea

Answer 22

hang on, lemme double check my last statement!

Answer 23

ok, that 8cos^4t is what is presenting an issue....I am looking at a few possible approaches....give me just a few more moments..don't want to lead you astray. But I usually work with the more complex side first.

Answer 24

ok, take your time

Answer 25

nvrmind...phi was correct in the approach.

Answer 26

you started with
\[\cos 4t = 8\cos ^{4}t-8\cos ^{2}t+1 \]

as you know, we can replace cos(4t) with
\[ 2 \cos ^{2}2t-1 \]

let's focus on cos(2t). using the same formula
\[ \cos(2t)=2\cos ^{2}t-1 \]
so we can replaces cos(2t) with that mess to get
\[ 2 \cos ^{2}2t-1 \text { becomes } 2(2\cos ^{2}t-1)^2 -1 \]

Answer 27

can you multiply out
\[ (2 \cos^2 t -1)^2 = (2 \cos^2 t -1)(2 \cos^2 t -1)\]
?

Answer 28

im looking

Answer 29

you can use FOIL (if you know that way)
it is the same as (x-y)(x-y)= x^2 -2 y + y^2
except x is 2 cos^2(t) and y is 1

Answer 30

this is crazy...

Answer 31

almost got it! wow, this is crazy! thank you both for your help! i do hope this question won;t be on my test on tuesday...

Answer 32

lol...anything is possible....not trying to scare you, btw..just need to be prepared for anything.

Answer 33

Here is how to do it using the distributive property
if the problem were (x+y)*A we would do x*A + y*A

let's call A= (2 cos^2t -1)
and multiply A times (2 cos^2t -1)
(2 cos^2t -1)A= 2 cos^2t * A - A
but A is (2 cos^2t -1), so this is

\[ (2 \cos^2t) (2 \cos^2t-1) - (2 \cos^2t -1)\]
now distribute again
\[ 2\cos^2t \cdot 2 \cos^2t - 2 \cos^2t - 2 \cos^2t -1 \]
or
\[ 4 \cos^4t - 4 \cos^2t -1 \]

Answer 34

Forgot the --1 at the end.

Answer 35

yes calmat noticed that should be +1 in
\[ 4 \cos^4t - 4 \cos^2t +1 \]

Answer 36

simplify
\[ 2 ( 4 \cos^4t - 4 \cos^2t +1) -1 \]

Answer 37

great explanation! Thank you both again!

Answer 38

to do this problem you had to use the cos(2x) --> 2 cos^2(x) -1
twice, to change cos(4x) into cos(x) (with exponents)