Use $cos 5theta$ to find the roots of $x(16x^4 - 20x^2 + 5) = 0$
$begingroup$
I used $cos(3theta + 2theta)$ to prove the first part, but I don't know how to the $2$nd part.
Show that $cos 5theta=16cos^5theta-20cos^3theta+5costheta,$ and hence show that
$$text{the roots of }x(16x^4 - 20x^2 + 5) text{ are: } 0,cosfrac{pi}{10}, cosfrac{3pi}{10},cosfrac{7pi}{10}, cosfrac{9pi}{10}$$
trigonometry complex-numbers
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add a comment |
$begingroup$
I used $cos(3theta + 2theta)$ to prove the first part, but I don't know how to the $2$nd part.
Show that $cos 5theta=16cos^5theta-20cos^3theta+5costheta,$ and hence show that
$$text{the roots of }x(16x^4 - 20x^2 + 5) text{ are: } 0,cosfrac{pi}{10}, cosfrac{3pi}{10},cosfrac{7pi}{10}, cosfrac{9pi}{10}$$
trigonometry complex-numbers
$endgroup$
add a comment |
$begingroup$
I used $cos(3theta + 2theta)$ to prove the first part, but I don't know how to the $2$nd part.
Show that $cos 5theta=16cos^5theta-20cos^3theta+5costheta,$ and hence show that
$$text{the roots of }x(16x^4 - 20x^2 + 5) text{ are: } 0,cosfrac{pi}{10}, cosfrac{3pi}{10},cosfrac{7pi}{10}, cosfrac{9pi}{10}$$
trigonometry complex-numbers
$endgroup$
I used $cos(3theta + 2theta)$ to prove the first part, but I don't know how to the $2$nd part.
Show that $cos 5theta=16cos^5theta-20cos^3theta+5costheta,$ and hence show that
$$text{the roots of }x(16x^4 - 20x^2 + 5) text{ are: } 0,cosfrac{pi}{10}, cosfrac{3pi}{10},cosfrac{7pi}{10}, cosfrac{9pi}{10}$$
trigonometry complex-numbers
trigonometry complex-numbers
edited Nov 25 '18 at 17:28
Lorenzo B.
1,8402520
1,8402520
asked Nov 25 '18 at 12:40
VanessaVanessa
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727
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2 Answers
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$begingroup$
Here the value of cosine will repeat after point.
Final values wil be only 5
$endgroup$
1
$begingroup$
Please use MathJax for writing answers. Because in MSE app its difficult to use image for some smartphones.
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– jayant98
Nov 25 '18 at 17:19
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I am using android phone, it is not showing option for Mathjax while writing answer
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– kapil pundir
Nov 25 '18 at 17:32
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math.meta.stackexchange.com/questions/5020/…
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– jayant98
Nov 25 '18 at 18:13
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Thankyou very much
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– kapil pundir
Nov 26 '18 at 15:59
add a comment |
$begingroup$
The equation $$;x(16x^4 - 20x^2 + 5) = 0; tag 1$$ can have $5$ real solutions (incl. multiplicity).
From $;cos 5theta = 16cos^5 theta - 20cos^3 theta + 5;$ we deduce that we will look for solutions of $(1)$ in the form of $cos theta;$ where $;cos 5theta = 0.$
$cos 5theta = 0$ holds for $10$ different values of $;theta in (-pi,pi];$ obtained when solving $$cos 5theta =frac pi2 +2kpiquad text{or} quad cos 5theta = -frac pi2 + 2kpi,; k=0,1,dots,4.$$
These are $$frac{pi}{10},; frac{5pi}{10},; frac{9pi}{10},;frac{13pi}{10},;frac{18pi}{10};text{and};-frac{pi}{10},;frac{3pi}{10},;frac{7pi}{10},;frac{11pi}{10},;frac{15pi}{10},;$$
between them two and two have equal cosines (can you check which are the pairs?).
The solutions are those listed in the question.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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active
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active
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votes
$begingroup$
Here the value of cosine will repeat after point.
Final values wil be only 5
$endgroup$
1
$begingroup$
Please use MathJax for writing answers. Because in MSE app its difficult to use image for some smartphones.
$endgroup$
– jayant98
Nov 25 '18 at 17:19
$begingroup$
I am using android phone, it is not showing option for Mathjax while writing answer
$endgroup$
– kapil pundir
Nov 25 '18 at 17:32
$begingroup$
math.meta.stackexchange.com/questions/5020/…
$endgroup$
– jayant98
Nov 25 '18 at 18:13
$begingroup$
Thankyou very much
$endgroup$
– kapil pundir
Nov 26 '18 at 15:59
add a comment |
$begingroup$
Here the value of cosine will repeat after point.
Final values wil be only 5
$endgroup$
1
$begingroup$
Please use MathJax for writing answers. Because in MSE app its difficult to use image for some smartphones.
$endgroup$
– jayant98
Nov 25 '18 at 17:19
$begingroup$
I am using android phone, it is not showing option for Mathjax while writing answer
$endgroup$
– kapil pundir
Nov 25 '18 at 17:32
$begingroup$
math.meta.stackexchange.com/questions/5020/…
$endgroup$
– jayant98
Nov 25 '18 at 18:13
$begingroup$
Thankyou very much
$endgroup$
– kapil pundir
Nov 26 '18 at 15:59
add a comment |
$begingroup$
Here the value of cosine will repeat after point.
Final values wil be only 5
$endgroup$
Here the value of cosine will repeat after point.
Final values wil be only 5
answered Nov 25 '18 at 16:56
kapil pundirkapil pundir
574
574
1
$begingroup$
Please use MathJax for writing answers. Because in MSE app its difficult to use image for some smartphones.
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– jayant98
Nov 25 '18 at 17:19
$begingroup$
I am using android phone, it is not showing option for Mathjax while writing answer
$endgroup$
– kapil pundir
Nov 25 '18 at 17:32
$begingroup$
math.meta.stackexchange.com/questions/5020/…
$endgroup$
– jayant98
Nov 25 '18 at 18:13
$begingroup$
Thankyou very much
$endgroup$
– kapil pundir
Nov 26 '18 at 15:59
add a comment |
1
$begingroup$
Please use MathJax for writing answers. Because in MSE app its difficult to use image for some smartphones.
$endgroup$
– jayant98
Nov 25 '18 at 17:19
$begingroup$
I am using android phone, it is not showing option for Mathjax while writing answer
$endgroup$
– kapil pundir
Nov 25 '18 at 17:32
$begingroup$
math.meta.stackexchange.com/questions/5020/…
$endgroup$
– jayant98
Nov 25 '18 at 18:13
$begingroup$
Thankyou very much
$endgroup$
– kapil pundir
Nov 26 '18 at 15:59
1
1
$begingroup$
Please use MathJax for writing answers. Because in MSE app its difficult to use image for some smartphones.
$endgroup$
– jayant98
Nov 25 '18 at 17:19
$begingroup$
Please use MathJax for writing answers. Because in MSE app its difficult to use image for some smartphones.
$endgroup$
– jayant98
Nov 25 '18 at 17:19
$begingroup$
I am using android phone, it is not showing option for Mathjax while writing answer
$endgroup$
– kapil pundir
Nov 25 '18 at 17:32
$begingroup$
I am using android phone, it is not showing option for Mathjax while writing answer
$endgroup$
– kapil pundir
Nov 25 '18 at 17:32
$begingroup$
math.meta.stackexchange.com/questions/5020/…
$endgroup$
– jayant98
Nov 25 '18 at 18:13
$begingroup$
math.meta.stackexchange.com/questions/5020/…
$endgroup$
– jayant98
Nov 25 '18 at 18:13
$begingroup$
Thankyou very much
$endgroup$
– kapil pundir
Nov 26 '18 at 15:59
$begingroup$
Thankyou very much
$endgroup$
– kapil pundir
Nov 26 '18 at 15:59
add a comment |
$begingroup$
The equation $$;x(16x^4 - 20x^2 + 5) = 0; tag 1$$ can have $5$ real solutions (incl. multiplicity).
From $;cos 5theta = 16cos^5 theta - 20cos^3 theta + 5;$ we deduce that we will look for solutions of $(1)$ in the form of $cos theta;$ where $;cos 5theta = 0.$
$cos 5theta = 0$ holds for $10$ different values of $;theta in (-pi,pi];$ obtained when solving $$cos 5theta =frac pi2 +2kpiquad text{or} quad cos 5theta = -frac pi2 + 2kpi,; k=0,1,dots,4.$$
These are $$frac{pi}{10},; frac{5pi}{10},; frac{9pi}{10},;frac{13pi}{10},;frac{18pi}{10};text{and};-frac{pi}{10},;frac{3pi}{10},;frac{7pi}{10},;frac{11pi}{10},;frac{15pi}{10},;$$
between them two and two have equal cosines (can you check which are the pairs?).
The solutions are those listed in the question.
$endgroup$
add a comment |
$begingroup$
The equation $$;x(16x^4 - 20x^2 + 5) = 0; tag 1$$ can have $5$ real solutions (incl. multiplicity).
From $;cos 5theta = 16cos^5 theta - 20cos^3 theta + 5;$ we deduce that we will look for solutions of $(1)$ in the form of $cos theta;$ where $;cos 5theta = 0.$
$cos 5theta = 0$ holds for $10$ different values of $;theta in (-pi,pi];$ obtained when solving $$cos 5theta =frac pi2 +2kpiquad text{or} quad cos 5theta = -frac pi2 + 2kpi,; k=0,1,dots,4.$$
These are $$frac{pi}{10},; frac{5pi}{10},; frac{9pi}{10},;frac{13pi}{10},;frac{18pi}{10};text{and};-frac{pi}{10},;frac{3pi}{10},;frac{7pi}{10},;frac{11pi}{10},;frac{15pi}{10},;$$
between them two and two have equal cosines (can you check which are the pairs?).
The solutions are those listed in the question.
$endgroup$
add a comment |
$begingroup$
The equation $$;x(16x^4 - 20x^2 + 5) = 0; tag 1$$ can have $5$ real solutions (incl. multiplicity).
From $;cos 5theta = 16cos^5 theta - 20cos^3 theta + 5;$ we deduce that we will look for solutions of $(1)$ in the form of $cos theta;$ where $;cos 5theta = 0.$
$cos 5theta = 0$ holds for $10$ different values of $;theta in (-pi,pi];$ obtained when solving $$cos 5theta =frac pi2 +2kpiquad text{or} quad cos 5theta = -frac pi2 + 2kpi,; k=0,1,dots,4.$$
These are $$frac{pi}{10},; frac{5pi}{10},; frac{9pi}{10},;frac{13pi}{10},;frac{18pi}{10};text{and};-frac{pi}{10},;frac{3pi}{10},;frac{7pi}{10},;frac{11pi}{10},;frac{15pi}{10},;$$
between them two and two have equal cosines (can you check which are the pairs?).
The solutions are those listed in the question.
$endgroup$
The equation $$;x(16x^4 - 20x^2 + 5) = 0; tag 1$$ can have $5$ real solutions (incl. multiplicity).
From $;cos 5theta = 16cos^5 theta - 20cos^3 theta + 5;$ we deduce that we will look for solutions of $(1)$ in the form of $cos theta;$ where $;cos 5theta = 0.$
$cos 5theta = 0$ holds for $10$ different values of $;theta in (-pi,pi];$ obtained when solving $$cos 5theta =frac pi2 +2kpiquad text{or} quad cos 5theta = -frac pi2 + 2kpi,; k=0,1,dots,4.$$
These are $$frac{pi}{10},; frac{5pi}{10},; frac{9pi}{10},;frac{13pi}{10},;frac{18pi}{10};text{and};-frac{pi}{10},;frac{3pi}{10},;frac{7pi}{10},;frac{11pi}{10},;frac{15pi}{10},;$$
between them two and two have equal cosines (can you check which are the pairs?).
The solutions are those listed in the question.
edited Nov 25 '18 at 17:07
amWhy
1
1
answered Nov 25 '18 at 15:21
user376343user376343
3,3582826
3,3582826
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