Zane Decker

2022-04-08

The quadratic equation whose roots are $\mathrm{sec}2\theta$ and $\mathrm{csc}\theta$ can be

a)${x}^{2}-2x+2=0$

b)${x}^{2}-5x+5=0$

c)${x}^{2}-7x+7=0$

d)${x}^{2}-9x+9=0$

a)

b)

c)

d)

Kendall Clark

Beginner2022-04-09Added 8 answers

Step 1

As others have pointed out, you’ve assume that if$u\ge v$ then $u}^{2}\ge {v}^{2$

This is true if u,v are both non-negative, but if both are non-positive, then the inequality reverses,$u}^{2}\le {v}^{2$ , and there is nothing we can say if u is positive and v negative.

Another way to write the equation is as

$(\left({\mathrm{cos}}^{2}\theta \right)x-1)(\left({\mathrm{sin}}^{2}\theta \right)x-1)=\left({\mathrm{cos}}^{2}\theta {\mathrm{sin}}^{2}\theta \right){x}^{2}-x+1$

So the monic polynomial with these two roots has

$p=q=\frac{1}{{\mathrm{sin}}^{2}\theta {\mathrm{cos}}^{2}\theta}=\frac{4}{{\mathrm{sin}}^{22}\theta}$

This is really the same as your first approach, but switches to sin and cos immediately.

As others have pointed out, you’ve assume that if

This is true if u,v are both non-negative, but if both are non-positive, then the inequality reverses,

Another way to write the equation is as

So the monic polynomial with these two roots has

This is really the same as your first approach, but switches to sin and cos immediately.

Kendall Wilkinson

Beginner2022-04-10Added 17 answers

Step 1

Recall that

$\frac{1}{{\mathrm{sec}}^{2}\theta}+\frac{1}{{\mathrm{csc}}^{2}\theta}=1.$ If

${x}^{2}+px+q=0$

then

$x={x}_{1}=\frac{-p+\sqrt{{p}^{2}-4q}}{2}$

or

$x={x}_{2}=\frac{-p-\sqrt{{p}^{2}-4q}}{2}.$

So you want$\frac{1}{{x}_{1}}+\frac{1}{{x}_{2}}=1$ You have

$\frac{1}{{x}_{1}}+\frac{1}{{x}_{2}}=\frac{2}{-p+\sqrt{{p}^{2}-4q}}+\frac{2}{-p-\sqrt{{p}^{2}-4q}}=\frac{-p}{q}$

So,

$1=\frac{-p}{q}$

$p=-q$

$\frac{-p\pm \sqrt{{p}^{2}+4p}}{2}=\{\begin{array}{ll}-1\pm \sqrt{3}& \text{if}p=2,\\ {\textstyle \phantom{\rule{1em}{0ex}}}\cdots & \text{if}p=5,\\ {\textstyle \phantom{\rule{1em}{0ex}}}\cdots & \text{if}p=7,\\ {\textstyle \phantom{\rule{1em}{0ex}}}\cdots & \text{if}p=9.\end{array}$

When$p=2$ , one of the solutions is negative and so cannot be $\mathrm{sec}2\theta$ or $\mathrm{csc}2\theta$ if $\theta$ is real. The other one is negative and so cannot be $\mathrm{sec}2\theta$ or $\mathrm{csc}2\theta$ if $\theta$ is real.

Go on from there is a similar way.

Recall that

then

or

So you want

So,

When

Go on from there is a similar way.

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