If $cos^4 alpha+4sin^4 beta-4sqrt{2}cos alpha sin beta +2=0$, then find $alpha$, $beta$ in $(0,fracpi2)$
$begingroup$
If $cos^4 alpha+4sin^4 beta-4sqrt{2}cos alpha sin beta +2=0$,
where $displaystyle alpha, beta in bigg(0,frac{pi}{2}bigg)$. Then value of $alpha,beta$ are
Try: I am trying to convert it into sum of square of quantity
like $$(cos^2 alpha)^2+(2sin^2 beta)^2-2cdot cos^2 alpha cdot 2sin^2 beta-4sqrt{2}cos alpha sin beta +2+4cos^2 alpha cdot sin^2 beta$$
$$(cos^2 alpha--2sin^2 beta)^2-4sqrt{2}cos alpha sin beta+4cos^2 alpha cdot sin^2 beta$$
Now i did not know how to solve it, could some help me
trigonometry
edited Jan 17 at 10:47
Blue
48.4k870154
asked Jan 17 at 10:01
DXTDXT
5,6892630
$endgroup$
add a comment |
$begingroup$
If $cos^4 alpha+4sin^4 beta-4sqrt{2}cos alpha sin beta +2=0$,
where $displaystyle alpha, beta in bigg(0,frac{pi}{2}bigg)$. Then value of $alpha,beta$ are
Try: I am trying to convert it into sum of square of quantity
like $$(cos^2 alpha)^2+(2sin^2 beta)^2-2cdot cos^2 alpha cdot 2sin^2 beta-4sqrt{2}cos alpha sin beta +2+4cos^2 alpha cdot sin^2 beta$$
$$(cos^2 alpha--2sin^2 beta)^2-4sqrt{2}cos alpha sin beta+4cos^2 alpha cdot sin^2 beta$$
Now i did not know how to solve it, could some help me
trigonometry
edited Jan 17 at 10:47
Blue
48.4k870154
asked Jan 17 at 10:01
DXTDXT
5,6892630
$endgroup$
$begingroup$
$cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
$endgroup$
– DXT
Jan 24 at 12:49
add a comment |
$begingroup$
If $cos^4 alpha+4sin^4 beta-4sqrt{2}cos alpha sin beta +2=0$,
where $displaystyle alpha, beta in bigg(0,frac{pi}{2}bigg)$. Then value of $alpha,beta$ are
Try: I am trying to convert it into sum of square of quantity
like $$(cos^2 alpha)^2+(2sin^2 beta)^2-2cdot cos^2 alpha cdot 2sin^2 beta-4sqrt{2}cos alpha sin beta +2+4cos^2 alpha cdot sin^2 beta$$
$$(cos^2 alpha--2sin^2 beta)^2-4sqrt{2}cos alpha sin beta+4cos^2 alpha cdot sin^2 beta$$
Now i did not know how to solve it, could some help me
trigonometry
edited Jan 17 at 10:47
Blue
48.4k870154
asked Jan 17 at 10:01
DXTDXT
5,6892630
$endgroup$
If $cos^4 alpha+4sin^4 beta-4sqrt{2}cos alpha sin beta +2=0$,
where $displaystyle alpha, beta in bigg(0,frac{pi}{2}bigg)$. Then value of $alpha,beta$ are
Try: I am trying to convert it into sum of square of quantity
like $$(cos^2 alpha)^2+(2sin^2 beta)^2-2cdot cos^2 alpha cdot 2sin^2 beta-4sqrt{2}cos alpha sin beta +2+4cos^2 alpha cdot sin^2 beta$$
$$(cos^2 alpha--2sin^2 beta)^2-4sqrt{2}cos alpha sin beta+4cos^2 alpha cdot sin^2 beta$$
Now i did not know how to solve it, could some help me
trigonometry
trigonometry
edited Jan 17 at 10:47
Blue
48.4k870154
asked Jan 17 at 10:01
DXTDXT
5,6892630
edited Jan 17 at 10:47
Blue
48.4k870154
asked Jan 17 at 10:01
DXTDXT
5,6892630
edited Jan 17 at 10:47
Blue
48.4k870154
edited Jan 17 at 10:47
Blue
48.4k870154
edited Jan 17 at 10:47
Blue
48.4k870154
48.4k870154
asked Jan 17 at 10:01
DXTDXT
5,6892630
asked Jan 17 at 10:01
DXTDXT
5,6892630
asked Jan 17 at 10:01
DXTDXT
5,6892630
5,6892630
$begingroup$
$cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
$endgroup$
– DXT
Jan 24 at 12:49
add a comment |
$begingroup$
$cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
$endgroup$
– DXT
Jan 24 at 12:49
$begingroup$
$cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
$endgroup$
– DXT
Jan 24 at 12:49
$begingroup$
$cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
$endgroup$
– DXT
Jan 24 at 12:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
For real $cos^2alpha,sin^2beta$
$$dfrac{cos^4alpha+4sin^4beta+1+1}4gesqrt[4]{cos^4alphacdot4sin^4beta}$$
The equality will occur if $cos^4alpha=4sin^4beta=1$
answered Jan 17 at 10:33
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
1
$begingroup$
See artofproblemsolving.com/wiki/index.php/…
$endgroup$
– lab bhattacharjee
Jan 17 at 11:01
add a comment |
$begingroup$
If you just call $x=cos(alpha)$ and $y=sin(beta)$ you get the 4th degree polynomial equation $x^4-4sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.
answered Jan 17 at 10:29
quaraguequarague
353110
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
For real $cos^2alpha,sin^2beta$
$$dfrac{cos^4alpha+4sin^4beta+1+1}4gesqrt[4]{cos^4alphacdot4sin^4beta}$$
The equality will occur if $cos^4alpha=4sin^4beta=1$
answered Jan 17 at 10:33
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
1
$begingroup$
See artofproblemsolving.com/wiki/index.php/…
$endgroup$
– lab bhattacharjee
Jan 17 at 11:01
add a comment |
$begingroup$
Hint:
For real $cos^2alpha,sin^2beta$
$$dfrac{cos^4alpha+4sin^4beta+1+1}4gesqrt[4]{cos^4alphacdot4sin^4beta}$$
The equality will occur if $cos^4alpha=4sin^4beta=1$
answered Jan 17 at 10:33
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
1
$begingroup$
See artofproblemsolving.com/wiki/index.php/…
$endgroup$
– lab bhattacharjee
Jan 17 at 11:01
add a comment |
$begingroup$
Hint:
For real $cos^2alpha,sin^2beta$
$$dfrac{cos^4alpha+4sin^4beta+1+1}4gesqrt[4]{cos^4alphacdot4sin^4beta}$$
The equality will occur if $cos^4alpha=4sin^4beta=1$
answered Jan 17 at 10:33
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
Hint:
For real $cos^2alpha,sin^2beta$
$$dfrac{cos^4alpha+4sin^4beta+1+1}4gesqrt[4]{cos^4alphacdot4sin^4beta}$$
The equality will occur if $cos^4alpha=4sin^4beta=1$
answered Jan 17 at 10:33
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 17 at 10:33
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 17 at 10:33
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 17 at 10:33
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
1
$begingroup$
See artofproblemsolving.com/wiki/index.php/…
$endgroup$
– lab bhattacharjee
Jan 17 at 11:01
add a comment |
1
$begingroup$
See artofproblemsolving.com/wiki/index.php/…
$endgroup$
– lab bhattacharjee
Jan 17 at 11:01
1
1
$begingroup$
See artofproblemsolving.com/wiki/index.php/…
$endgroup$
– lab bhattacharjee
Jan 17 at 11:01
$begingroup$
See artofproblemsolving.com/wiki/index.php/…
$endgroup$
– lab bhattacharjee
Jan 17 at 11:01
add a comment |
$begingroup$
If you just call $x=cos(alpha)$ and $y=sin(beta)$ you get the 4th degree polynomial equation $x^4-4sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.
answered Jan 17 at 10:29
quaraguequarague
353110
$endgroup$
add a comment |
$begingroup$
If you just call $x=cos(alpha)$ and $y=sin(beta)$ you get the 4th degree polynomial equation $x^4-4sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.
answered Jan 17 at 10:29
quaraguequarague
353110
$endgroup$
add a comment |
$begingroup$
If you just call $x=cos(alpha)$ and $y=sin(beta)$ you get the 4th degree polynomial equation $x^4-4sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.
answered Jan 17 at 10:29
quaraguequarague
353110
$endgroup$
If you just call $x=cos(alpha)$ and $y=sin(beta)$ you get the 4th degree polynomial equation $x^4-4sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.
answered Jan 17 at 10:29
quaraguequarague
353110
answered Jan 17 at 10:29
quaraguequarague
353110
answered Jan 17 at 10:29
quaraguequarague
353110
answered Jan 17 at 10:29
quaraguequarague
353110
353110
add a comment |
add a comment |
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$begingroup$
$cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
$endgroup$
– DXT
Jan 24 at 12:49