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Prove that $4cos^4x-2cos2x-1/2cos4x$ is independent of x
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$4cos^4x-2cos2x-1/2cos4x$
I don't know how to proceed. $cos4x=2cos^2(2x)-1$ Is this useful at all?
trigonometry
asked Aug 6 at 16:22
James Warthington
21227
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up vote
1
down vote
favorite
$4cos^4x-2cos2x-1/2cos4x$
I don't know how to proceed. $cos4x=2cos^2(2x)-1$ Is this useful at all?
trigonometry
asked Aug 6 at 16:22
James Warthington
21227
The simplest way is to exploit $2cos z=e^iz+e^-iz$, but you may also check that $int_0^2pif'(x)^2,dx = 0$, so $f$ is constant.
– Jack D'Aurizio♦
Aug 6 at 17:40
add a comment |Â
up vote
1
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favorite
up vote
1
down vote
favorite
$4cos^4x-2cos2x-1/2cos4x$
I don't know how to proceed. $cos4x=2cos^2(2x)-1$ Is this useful at all?
trigonometry
asked Aug 6 at 16:22
James Warthington
21227
$4cos^4x-2cos2x-1/2cos4x$
I don't know how to proceed. $cos4x=2cos^2(2x)-1$ Is this useful at all?
trigonometry
asked Aug 6 at 16:22
James Warthington
21227
asked Aug 6 at 16:22
James Warthington
21227
asked Aug 6 at 16:22
James Warthington
21227
asked Aug 6 at 16:22
James Warthington
21227
21227
The simplest way is to exploit $2cos z=e^iz+e^-iz$, but you may also check that $int_0^2pif'(x)^2,dx = 0$, so $f$ is constant.
– Jack D'Aurizio♦
Aug 6 at 17:40
add a comment |Â
The simplest way is to exploit $2cos z=e^iz+e^-iz$, but you may also check that $int_0^2pif'(x)^2,dx = 0$, so $f$ is constant.
– Jack D'Aurizio♦
Aug 6 at 17:40
The simplest way is to exploit $2cos z=e^iz+e^-iz$, but you may also check that $int_0^2pif'(x)^2,dx = 0$, so $f$ is constant.
– Jack D'Aurizio♦
Aug 6 at 17:40
The simplest way is to exploit $2cos z=e^iz+e^-iz$, but you may also check that $int_0^2pif'(x)^2,dx = 0$, so $f$ is constant.
– Jack D'Aurizio♦
Aug 6 at 17:40
add a comment |Â
4 Answers
4
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oldest
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up vote
1
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accepted
Yes, $cos4x=2cos^2(2x)-1$ is useful, also $cos^2x=dfrac1+cos2x2$.
Now
$$4cos^4x-2cos2x-1/2cos4x=4(dfrac1+cos2x2)^2-2cos2x-1/2(2cos^2(2x)-1)$$
then let $cos2x=k$. Can you proceed?
answered Aug 6 at 16:52
Nosrati
20.1k41644
add a comment |Â
up vote
1
down vote
$$4cos^4x-2cos2x-frac12cos4x=(1+cos2x)^2-2cos2x-frac12(2cos^22x-1)=frac32.$$
answered Aug 6 at 16:57
Michael Rozenberg
88.2k1579180
How $4cos^4(x)=(1-cos2x)^2$? Isn't the latter equivalent to $1-2cos2x+cos^2(2x)$
– James Warthington
Aug 6 at 17:45
I said $4cos^4x=(1+cos2x)^2.$ Because $(1+cos2x)^2=(1+2cos^2x-1)^2=4cos^4x$.
– Michael Rozenberg
Aug 6 at 19:13
Why someone wants to delete my solution? Explain please your step?
– Michael Rozenberg
Aug 7 at 8:49
add a comment |Â
up vote
0
down vote
Hint: using $cos2x = 1-2sin^2(x)$, $sin(2x) = 2sin(x)cos(x)$ and $sin^2(x) + cos^2(x) = 1$ to get the following and simplify it:
$$S = 4cos^4(x)-2(1-2sin^2(x))+1/2(1-2sin^2(2x))$$
$$S = 4cos^4(x)-2+4sin^2(x)+1/2 - 4sin^2(x)cos^2(x)$$
$$S = 4cos^4(x)-2+1/2 + 4sin^2(x)(1-cos^2(x)) = 4cos^4(x)-2+1/2 + 4sin^4(x) = 4 - 2 + 1/2 = 2.5$$
I should notice that I think one of the $-$ is $+$ (in my case the second minus).
answered Aug 6 at 16:31
OmG
1,740617
add a comment |Â
up vote
0
down vote
Yes but an alternative is to use the definition in terms of exponentials
$$
cos x=frac12(e^ix+e^-ix)
$$
So
$$
cos^4 x=frac12^4(e^ix+e^-ix)^4
$$
If you expand the right-hand side with the binomial theorem and re-apply the exponential definition of $cos$ e.g. $cos 2x = frac12(e^i2x+e^-i2x)$ the result appears.
answered Aug 6 at 16:56
PM.
3,1052822
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, $cos4x=2cos^2(2x)-1$ is useful, also $cos^2x=dfrac1+cos2x2$.
Now
$$4cos^4x-2cos2x-1/2cos4x=4(dfrac1+cos2x2)^2-2cos2x-1/2(2cos^2(2x)-1)$$
then let $cos2x=k$. Can you proceed?
answered Aug 6 at 16:52
Nosrati
20.1k41644
add a comment |Â
up vote
1
down vote
accepted
Yes, $cos4x=2cos^2(2x)-1$ is useful, also $cos^2x=dfrac1+cos2x2$.
Now
$$4cos^4x-2cos2x-1/2cos4x=4(dfrac1+cos2x2)^2-2cos2x-1/2(2cos^2(2x)-1)$$
then let $cos2x=k$. Can you proceed?
answered Aug 6 at 16:52
Nosrati
20.1k41644
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, $cos4x=2cos^2(2x)-1$ is useful, also $cos^2x=dfrac1+cos2x2$.
Now
$$4cos^4x-2cos2x-1/2cos4x=4(dfrac1+cos2x2)^2-2cos2x-1/2(2cos^2(2x)-1)$$
then let $cos2x=k$. Can you proceed?
answered Aug 6 at 16:52
Nosrati
20.1k41644
Yes, $cos4x=2cos^2(2x)-1$ is useful, also $cos^2x=dfrac1+cos2x2$.
Now
$$4cos^4x-2cos2x-1/2cos4x=4(dfrac1+cos2x2)^2-2cos2x-1/2(2cos^2(2x)-1)$$
then let $cos2x=k$. Can you proceed?
answered Aug 6 at 16:52
Nosrati
20.1k41644
answered Aug 6 at 16:52
Nosrati
20.1k41644
answered Aug 6 at 16:52
Nosrati
20.1k41644
answered Aug 6 at 16:52
Nosrati
20.1k41644
20.1k41644
add a comment |Â
add a comment |Â
up vote
1
down vote
$$4cos^4x-2cos2x-frac12cos4x=(1+cos2x)^2-2cos2x-frac12(2cos^22x-1)=frac32.$$
answered Aug 6 at 16:57
Michael Rozenberg
88.2k1579180
How $4cos^4(x)=(1-cos2x)^2$? Isn't the latter equivalent to $1-2cos2x+cos^2(2x)$
– James Warthington
Aug 6 at 17:45
I said $4cos^4x=(1+cos2x)^2.$ Because $(1+cos2x)^2=(1+2cos^2x-1)^2=4cos^4x$.
– Michael Rozenberg
Aug 6 at 19:13
Why someone wants to delete my solution? Explain please your step?
– Michael Rozenberg
Aug 7 at 8:49
add a comment |Â
up vote
1
down vote
$$4cos^4x-2cos2x-frac12cos4x=(1+cos2x)^2-2cos2x-frac12(2cos^22x-1)=frac32.$$
answered Aug 6 at 16:57
Michael Rozenberg
88.2k1579180
How $4cos^4(x)=(1-cos2x)^2$? Isn't the latter equivalent to $1-2cos2x+cos^2(2x)$
– James Warthington
Aug 6 at 17:45
I said $4cos^4x=(1+cos2x)^2.$ Because $(1+cos2x)^2=(1+2cos^2x-1)^2=4cos^4x$.
– Michael Rozenberg
Aug 6 at 19:13
Why someone wants to delete my solution? Explain please your step?
– Michael Rozenberg
Aug 7 at 8:49
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$4cos^4x-2cos2x-frac12cos4x=(1+cos2x)^2-2cos2x-frac12(2cos^22x-1)=frac32.$$
answered Aug 6 at 16:57
Michael Rozenberg
88.2k1579180
$$4cos^4x-2cos2x-frac12cos4x=(1+cos2x)^2-2cos2x-frac12(2cos^22x-1)=frac32.$$
answered Aug 6 at 16:57
Michael Rozenberg
88.2k1579180
answered Aug 6 at 16:57
Michael Rozenberg
88.2k1579180
answered Aug 6 at 16:57
Michael Rozenberg
88.2k1579180
answered Aug 6 at 16:57
Michael Rozenberg
88.2k1579180
88.2k1579180
How $4cos^4(x)=(1-cos2x)^2$? Isn't the latter equivalent to $1-2cos2x+cos^2(2x)$
– James Warthington
Aug 6 at 17:45
I said $4cos^4x=(1+cos2x)^2.$ Because $(1+cos2x)^2=(1+2cos^2x-1)^2=4cos^4x$.
– Michael Rozenberg
Aug 6 at 19:13
Why someone wants to delete my solution? Explain please your step?
– Michael Rozenberg
Aug 7 at 8:49
add a comment |Â
How $4cos^4(x)=(1-cos2x)^2$? Isn't the latter equivalent to $1-2cos2x+cos^2(2x)$
– James Warthington
Aug 6 at 17:45
I said $4cos^4x=(1+cos2x)^2.$ Because $(1+cos2x)^2=(1+2cos^2x-1)^2=4cos^4x$.
– Michael Rozenberg
Aug 6 at 19:13
Why someone wants to delete my solution? Explain please your step?
– Michael Rozenberg
Aug 7 at 8:49
How $4cos^4(x)=(1-cos2x)^2$? Isn't the latter equivalent to $1-2cos2x+cos^2(2x)$
– James Warthington
Aug 6 at 17:45
How $4cos^4(x)=(1-cos2x)^2$? Isn't the latter equivalent to $1-2cos2x+cos^2(2x)$
– James Warthington
Aug 6 at 17:45
I said $4cos^4x=(1+cos2x)^2.$ Because $(1+cos2x)^2=(1+2cos^2x-1)^2=4cos^4x$.
– Michael Rozenberg
Aug 6 at 19:13
I said $4cos^4x=(1+cos2x)^2.$ Because $(1+cos2x)^2=(1+2cos^2x-1)^2=4cos^4x$.
– Michael Rozenberg
Aug 6 at 19:13
Why someone wants to delete my solution? Explain please your step?
– Michael Rozenberg
Aug 7 at 8:49
Why someone wants to delete my solution? Explain please your step?
– Michael Rozenberg
Aug 7 at 8:49
add a comment |Â
up vote
0
down vote
Hint: using $cos2x = 1-2sin^2(x)$, $sin(2x) = 2sin(x)cos(x)$ and $sin^2(x) + cos^2(x) = 1$ to get the following and simplify it:
$$S = 4cos^4(x)-2(1-2sin^2(x))+1/2(1-2sin^2(2x))$$
$$S = 4cos^4(x)-2+4sin^2(x)+1/2 - 4sin^2(x)cos^2(x)$$
$$S = 4cos^4(x)-2+1/2 + 4sin^2(x)(1-cos^2(x)) = 4cos^4(x)-2+1/2 + 4sin^4(x) = 4 - 2 + 1/2 = 2.5$$
I should notice that I think one of the $-$ is $+$ (in my case the second minus).
answered Aug 6 at 16:31
OmG
1,740617
add a comment |Â
up vote
0
down vote
Hint: using $cos2x = 1-2sin^2(x)$, $sin(2x) = 2sin(x)cos(x)$ and $sin^2(x) + cos^2(x) = 1$ to get the following and simplify it:
$$S = 4cos^4(x)-2(1-2sin^2(x))+1/2(1-2sin^2(2x))$$
$$S = 4cos^4(x)-2+4sin^2(x)+1/2 - 4sin^2(x)cos^2(x)$$
$$S = 4cos^4(x)-2+1/2 + 4sin^2(x)(1-cos^2(x)) = 4cos^4(x)-2+1/2 + 4sin^4(x) = 4 - 2 + 1/2 = 2.5$$
I should notice that I think one of the $-$ is $+$ (in my case the second minus).
answered Aug 6 at 16:31
OmG
1,740617
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: using $cos2x = 1-2sin^2(x)$, $sin(2x) = 2sin(x)cos(x)$ and $sin^2(x) + cos^2(x) = 1$ to get the following and simplify it:
$$S = 4cos^4(x)-2(1-2sin^2(x))+1/2(1-2sin^2(2x))$$
$$S = 4cos^4(x)-2+4sin^2(x)+1/2 - 4sin^2(x)cos^2(x)$$
$$S = 4cos^4(x)-2+1/2 + 4sin^2(x)(1-cos^2(x)) = 4cos^4(x)-2+1/2 + 4sin^4(x) = 4 - 2 + 1/2 = 2.5$$
I should notice that I think one of the $-$ is $+$ (in my case the second minus).
answered Aug 6 at 16:31
OmG
1,740617
Hint: using $cos2x = 1-2sin^2(x)$, $sin(2x) = 2sin(x)cos(x)$ and $sin^2(x) + cos^2(x) = 1$ to get the following and simplify it:
$$S = 4cos^4(x)-2(1-2sin^2(x))+1/2(1-2sin^2(2x))$$
$$S = 4cos^4(x)-2+4sin^2(x)+1/2 - 4sin^2(x)cos^2(x)$$
$$S = 4cos^4(x)-2+1/2 + 4sin^2(x)(1-cos^2(x)) = 4cos^4(x)-2+1/2 + 4sin^4(x) = 4 - 2 + 1/2 = 2.5$$
I should notice that I think one of the $-$ is $+$ (in my case the second minus).
answered Aug 6 at 16:31
OmG
1,740617
edited Aug 6 at 16:41
answered Aug 6 at 16:31
OmG
1,740617
answered Aug 6 at 16:31
OmG
1,740617
answered Aug 6 at 16:31
OmG
1,740617
1,740617
add a comment |Â
add a comment |Â
up vote
0
down vote
Yes but an alternative is to use the definition in terms of exponentials
$$
cos x=frac12(e^ix+e^-ix)
$$
So
$$
cos^4 x=frac12^4(e^ix+e^-ix)^4
$$
If you expand the right-hand side with the binomial theorem and re-apply the exponential definition of $cos$ e.g. $cos 2x = frac12(e^i2x+e^-i2x)$ the result appears.
answered Aug 6 at 16:56
PM.
3,1052822
add a comment |Â
up vote
0
down vote
Yes but an alternative is to use the definition in terms of exponentials
$$
cos x=frac12(e^ix+e^-ix)
$$
So
$$
cos^4 x=frac12^4(e^ix+e^-ix)^4
$$
If you expand the right-hand side with the binomial theorem and re-apply the exponential definition of $cos$ e.g. $cos 2x = frac12(e^i2x+e^-i2x)$ the result appears.
answered Aug 6 at 16:56
PM.
3,1052822
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes but an alternative is to use the definition in terms of exponentials
$$
cos x=frac12(e^ix+e^-ix)
$$
So
$$
cos^4 x=frac12^4(e^ix+e^-ix)^4
$$
If you expand the right-hand side with the binomial theorem and re-apply the exponential definition of $cos$ e.g. $cos 2x = frac12(e^i2x+e^-i2x)$ the result appears.
answered Aug 6 at 16:56
PM.
3,1052822
Yes but an alternative is to use the definition in terms of exponentials
$$
cos x=frac12(e^ix+e^-ix)
$$
So
$$
cos^4 x=frac12^4(e^ix+e^-ix)^4
$$
If you expand the right-hand side with the binomial theorem and re-apply the exponential definition of $cos$ e.g. $cos 2x = frac12(e^i2x+e^-i2x)$ the result appears.
answered Aug 6 at 16:56
PM.
3,1052822
answered Aug 6 at 16:56
PM.
3,1052822
answered Aug 6 at 16:56
PM.
3,1052822
answered Aug 6 at 16:56
PM.
3,1052822
3,1052822
add a comment |Â
add a comment |Â
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The simplest way is to exploit $2cos z=e^iz+e^-iz$, but you may also check that $int_0^2pif'(x)^2,dx = 0$, so $f$ is constant.
– Jack D'Aurizio♦
Aug 6 at 17:40