Trigonometric equation with 2 variables
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Find all real values of $x$ and $y$ such that $sin^4x+cos^4y+2=4sin xcos y$ .
I started with $u=sin x$, $v=cos y$
I then can show that the above expression can be written as
$(u^2-1)+(v^2-1)+2 (u-v)^2=0$
After this I am unable to find the values of $u$ and $v$.
Please can anyone help me?
Thank you
trigonometry
edited Nov 18 at 7:47
Andrei
10.3k21025
asked Nov 18 at 7:42
Ashwini
62
add a comment |
up vote
0
down vote
favorite
Find all real values of $x$ and $y$ such that $sin^4x+cos^4y+2=4sin xcos y$ .
I started with $u=sin x$, $v=cos y$
I then can show that the above expression can be written as
$(u^2-1)+(v^2-1)+2 (u-v)^2=0$
After this I am unable to find the values of $u$ and $v$.
Please can anyone help me?
Thank you
trigonometry
edited Nov 18 at 7:47
Andrei
10.3k21025
asked Nov 18 at 7:42
Ashwini
62
It should be $left(u^2-1right)^2+left(v^2-1right)^2+2(u-v)^2=0$.
– Lozenges
Nov 18 at 8:09
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find all real values of $x$ and $y$ such that $sin^4x+cos^4y+2=4sin xcos y$ .
I started with $u=sin x$, $v=cos y$
I then can show that the above expression can be written as
$(u^2-1)+(v^2-1)+2 (u-v)^2=0$
After this I am unable to find the values of $u$ and $v$.
Please can anyone help me?
Thank you
trigonometry
edited Nov 18 at 7:47
Andrei
10.3k21025
asked Nov 18 at 7:42
Ashwini
62
Find all real values of $x$ and $y$ such that $sin^4x+cos^4y+2=4sin xcos y$ .
I started with $u=sin x$, $v=cos y$
I then can show that the above expression can be written as
$(u^2-1)+(v^2-1)+2 (u-v)^2=0$
After this I am unable to find the values of $u$ and $v$.
Please can anyone help me?
Thank you
trigonometry
trigonometry
edited Nov 18 at 7:47
Andrei
10.3k21025
asked Nov 18 at 7:42
Ashwini
62
edited Nov 18 at 7:47
Andrei
10.3k21025
asked Nov 18 at 7:42
Ashwini
62
edited Nov 18 at 7:47
Andrei
10.3k21025
edited Nov 18 at 7:47
Andrei
10.3k21025
edited Nov 18 at 7:47
Andrei
10.3k21025
10.3k21025
asked Nov 18 at 7:42
Ashwini
62
asked Nov 18 at 7:42
Ashwini
62
asked Nov 18 at 7:42
Ashwini
62
62
It should be $left(u^2-1right)^2+left(v^2-1right)^2+2(u-v)^2=0$.
– Lozenges
Nov 18 at 8:09
add a comment |
It should be $left(u^2-1right)^2+left(v^2-1right)^2+2(u-v)^2=0$.
– Lozenges
Nov 18 at 8:09
It should be $left(u^2-1right)^2+left(v^2-1right)^2+2(u-v)^2=0$.
– Lozenges
Nov 18 at 8:09
It should be $left(u^2-1right)^2+left(v^2-1right)^2+2(u-v)^2=0$.
– Lozenges
Nov 18 at 8:09
add a comment |
1 Answer
1
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With the correction pointed out by @Lozenges, you have a sum of square numbers that is equal to $0$. That is true only if all terms are $0$. The solution is then given by $$sin x=cos y=pm 1$$
You can write $$begin{align}x&=frac{pi}2+npi\y&=npiend{align}$$ with $ninmathbb Z$.
answered Nov 18 at 8:13
Andrei
10.3k21025
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
With the correction pointed out by @Lozenges, you have a sum of square numbers that is equal to $0$. That is true only if all terms are $0$. The solution is then given by $$sin x=cos y=pm 1$$
You can write $$begin{align}x&=frac{pi}2+npi\y&=npiend{align}$$ with $ninmathbb Z$.
answered Nov 18 at 8:13
Andrei
10.3k21025
add a comment |
up vote
0
down vote
With the correction pointed out by @Lozenges, you have a sum of square numbers that is equal to $0$. That is true only if all terms are $0$. The solution is then given by $$sin x=cos y=pm 1$$
You can write $$begin{align}x&=frac{pi}2+npi\y&=npiend{align}$$ with $ninmathbb Z$.
answered Nov 18 at 8:13
Andrei
10.3k21025
add a comment |
up vote
0
down vote
up vote
0
down vote
With the correction pointed out by @Lozenges, you have a sum of square numbers that is equal to $0$. That is true only if all terms are $0$. The solution is then given by $$sin x=cos y=pm 1$$
You can write $$begin{align}x&=frac{pi}2+npi\y&=npiend{align}$$ with $ninmathbb Z$.
answered Nov 18 at 8:13
Andrei
10.3k21025
With the correction pointed out by @Lozenges, you have a sum of square numbers that is equal to $0$. That is true only if all terms are $0$. The solution is then given by $$sin x=cos y=pm 1$$
You can write $$begin{align}x&=frac{pi}2+npi\y&=npiend{align}$$ with $ninmathbb Z$.
answered Nov 18 at 8:13
Andrei
10.3k21025
answered Nov 18 at 8:13
Andrei
10.3k21025
answered Nov 18 at 8:13
Andrei
10.3k21025
answered Nov 18 at 8:13
Andrei
10.3k21025
10.3k21025
add a comment |
add a comment |
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It should be $left(u^2-1right)^2+left(v^2-1right)^2+2(u-v)^2=0$.
– Lozenges
Nov 18 at 8:09