Question

Prove the Identity.

cos(4x)=cos^(4)x-6sin^(2)xcos^(2)x+sin^(4)x

Answer 1

\[\cos^{4}x-6\sin^{2}xcos^{2}x+\sin^{4}x\]
\[= .5(1+\cos^{2}(2x)) -6*0.5(1-\cos{2x})*0.5(1+\cos{2x}) +0.5(1-\cos{2x}) \]
\[=.5+(.5\cos^{2}{2x})-3*0.5+(3*0.5\cos{2x})-3*0.5-(3*0.5\cos{2x})\]
\[=-6.5+0.5\cos^{2}{2x}\]
\[=-6.5+\frac{1}{4}(1+\cos4x)\]

This is all i know .. it maybe further simplified .
regards

Answer 2

I have replaced 1 with:

\[\large \sin^2(4x) + \cos^2(4x) = 1\]

Answer 3

And somehow I am getting -8 and not -6 in the middle @rosiepose99

Check your question once again..

Answer 4

I think I have done something wrong..

Answer 5

it is -6

Answer 6

\(\cos(4x) = \cos2(2x) \implies 1 - 2\sin^2(2x) \implies 1 - 2(\sin(2x)^2)\)

\(= 1 - 2(2\sin(x)\cos(x))^2 \implies 1 - 2(4\sin^2(x) \cos^2(x)) \implies1 - 8\sin^2(x) \cdot(\cos^2(x))\)

Now using:

\[\cos^4(x) + \sin^4(x) = 1 - 2\sin^2(x) \cdot \cos^2(x)\]

\[\implies 1 = \cos^4(x) + \sin^4(x) + 2\sin^2(x) \cdot \cos^2(x)\]

Replace 1 with it:

\[\implies1 - 8\sin^2(x) \cdot(\cos^2(x))\]

\[\implies \cos^4(x) + \sin^4(x) + 2\sin^2(x) \cdot \cos^2(x) - 8\sin^2(x) \cdot \cos^2(x)\]

\[ \implies \color{green}{\cos^4(x) + \sin^4(x) - 6\sin^2(x) \cdot \cos^2(x)}\]

Answer 7

where you got cos^4x +sin^4x =1-2sin^2c*cos^2x
why?

Answer 8

Do you know :

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

@rosiepose99

Answer 9

And do you know:

\[(a + b)^2 = a^2 + b^2 + 2ab\]

Answer 10

here put \(a = cos^2{x}\) and b = \(sin^2(x)\)

Answer 11

ok

Answer 12

We will get:

\[(\cos^2(x) + \sin^2(x))^2 = (\cos^2(x))^2 + (\sin^2(x))^2 + 2\sin^2(x)\cos^2(x)\]

\[(\cos^2(x) + \sin^2(x))^2 = \cos^4(x) + \sin^4(x) + 2\sin^2(x)\cos^2(x)\]

As:
\(\cos^2(x) + \sin^2(x) = 1\) put this on LHS:

\[\implies 1 = \cos^4(x) + \sin^4(x) + 2\sin^2(x)\cos^2(x)\]

Getting @rosiepose99

Answer 13

yes, but where you replace sin^2x +cos^2x=1 you would be left with another sin^2x+cos^2x right...and it shows sin^4x +cos^4x??????

Answer 14

I replace on left hand side..

Answer 15

See:

\[(\cos^2(x) + \sin^2(x))^2 = \cos^4(x) + \sin^4(x) + 2\sin^2(x)\cos^2(x)\]

\[(1)^2 = \cos^4(x) + \sin^4(x) + 2\sin^2(x)\cos^2(x)\]

Answer 16

Getting now @rosiepose99

put this value in place of 1 there..

Answer 17

I will show you where I have replaced.. Just look for colored one:

\[(\color{green}{\cos^2(x) + \sin^2(x)})^2 = \cos^4(x) + \sin^4(x) + 2\sin^2(x)\cos^2(x)\]

Answer 18

Do reply @rosiepose99

Answer 19

ok but it looks like u r using cos^2x +sin^2x twice

Answer 20

Where??

Answer 21

one sec.

Answer 22

See there is difference between:

\[\large (\cos^2(x) + \sin^2(x))^2\]

and :

\[\large (\cos^2(x))^2 + (\sin^2(x))^2\]

Answer 23

First one is:

\[\large (1)^2 \implies 1\]

Answer 24

And the second one is:

\[\large \cos^4(x) + \sin^4(x)\]

Answer 25

cos^4(x)+sin^4(x)=1−2sin^2(x)⋅cos^2(x)

this is where i am confused.....?y?

Answer 26

See I have used the above one in my way.

So can you forget this?/

Just apply the identity that I have just given that:

\[(1)^2 = \cos^4(x) + \sin^4(x) + 2\sin^2(x)\cos^2(x)\]

Answer 27

ok

Answer 28

See I don't want you to cram it..

I will surely show you how I did..

Answer 29

Just reply fast @rosiepose99
Okay??

Answer 30

ok

Answer 31

See:

\[(a+b)^2 = a^2 + b^2 + 2ab\]

Right??

Answer 32

yes

Answer 33

Subtract both the sides by 2ab:

\[(a+b)^2 - 2ab = a^2 + b^2 + 2ab - 2ab\]

Right??

Answer 34

ok

Answer 35

Didn't get??

Answer 36

So 2ab and -2ab on right hand side will get cancel:
\[(a+b)^2 - 2ab = a^2 + b^2 + \cancel{2ab} - \cancel{2ab}\]

Rght??

Answer 37

yes

Answer 38

So you are left with:

\[a^2 + b^2 = (a+b)^2 - 2ab\]

right??

I have just exchanged RHS and LHS..

Answer 39

k

Answer 40

Put a = \(cos^2(x)\) and b = \(sin^2(x)\) you will get:

\[(\cos^2(x))^2 + (\sin^2(x))^2 = (\sin^2(x) + \cos^2(x)) - 2\sin^2(x) \cdot \cos^2(x)\]

Right ??

Look carefully..

Answer 41

Sorry I did wrong..

Answer 42

y did square twice

Answer 43

Wait..

Answer 44

oh

Answer 45

k

Answer 46

\[(\cos^2(x))^2 + ((\sin^2(x))^2 = (\sin^2(x) + \cos^2(x))^2 - 2\sin^2(x) \cdot \cos^2(x)\]

Just same as:

\[a^2 + b^2 = (a+b)^2 - 2ab\]

Answer 47

So firstly solve LHS:

\[(\cos^2(x))^2 + (\sin^2(x))^2 = \cos^4(x) + \sin^4(x)\]

Right?/

this is LHS..

Answer 48

yes

Answer 49

can u show/explain to me how/why you got cos4x=1-8sin^2xcos^2x

Answer 50

Now solve RHS:

\[(\sin^2(x) + \cos^2(x))^2 - 2\sin^2(x) \cdot \cos^2(x) \implies (1)^2 - 2\sin^2(x) \cdot \cos^2(x)\]

Answer 51

on the LHS

Answer 52

Do you know:

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

Answer 53

no, is that the double angle identity?

Answer 54

yeah it is, ok

Answer 55

They are half angle and double identities..

You have to learn them..

Answer 56

yes i remember

Answer 57

Double angle..

Answer 58

In LHS whatever be the angle on RHS it becomes half..

Okay??

Answer 59

k

Answer 60

So can I write:

\(cos(4x)\) as \(cos2(2x)\)

tell me??

Answer 61

yes

Answer 62

So:

\[\cos(4x) = 1 - 2\sin^2(2x)\]

Right??

Answer 63

yes

Answer 64

but then you squared,,,y

Answer 65

Where??

Answer 66

I am showing you step by step so be cool..

Last step you got??

Answer 67

\[\cos(4x) = 1 - 2\sin^2(2x)\]

Go till here??

Answer 68

ok, i misread because you rewrote the power to the end of the parenthesis.

Answer 69

Yes...

Answer 70

See:
\(sin^2(x)\) can be written as : \((sin(x))^2\)

Okay??

Answer 71

Do you know this:

\[\sin(2x) = 2 \sin(x)\cos(x)\]

Answer 72

no i didn't,

Answer 73

Remember this always..

Answer 74

It is also double and half angle identity...

Answer 75

k

Answer 76

Do you want its proof or you will remember this??

Answer 77

I just looked at notes and see the identity, thanks anyway

Answer 78

\[1 - 2(\sin(2x)^2) \implies 1 - 2(2\sin(x)\cos(x))^2\]

Right??

Answer 79

yes

Answer 80

Square the last:

\[1 - 2(2\sin(x)\cos(x))^2 \implies 1 - 2(4\sin^2(x)\cos^2(x))\]

Right??

Answer 81

yes

Answer 82

Now multiply 2 with 4:
\[1 - 2(4\sin^2(x)\cos^2(x)) \implies 1 - 8\sin^2(x)\cos^2(x)\]

Answer 83

Got all the steps??

Form now replace 1 with that I told you above..


Got??

Answer 84

k

Answer 85

Go through all the steps you will definitely get that..

Answer 86

ok, back to the one you explained first we r replacing 1, but where did you get the cos^4x +sin^4x. Is it from the other side?

Answer 87

why did you choose to cos^4x + sin^4x