hadejada7x
2022-01-15
Answered

Given $\mathrm{cos}\left(a\right)+\mathrm{cos}\left(b\right)=1$ , prove that $1-{s}^{2}-{t}^{2}-3{s}^{2}{t}^{2}=0$ , where $s=\mathrm{tan}\left(\frac{a}{2}\right)$ and $t=\mathrm{tan}\left(\frac{b}{2}\right)$

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Elaine Verrett

Answered 2022-01-16
Author has **41** answers

Alternatively:

$\mathrm{cos}a=2{\mathrm{cos}}^{2}\frac{a}{2}-1$

$\mathrm{cos}}^{2}\frac{a}{2}=\frac{1}{1+{\mathrm{tan}}^{2}\frac{a}{2}}=\frac{1}{1+{s}^{2}$

$\mathrm{cos}a+\mathrm{cos}b=2{\mathrm{cos}}^{2}\frac{a}{2}+2{\mathrm{cos}}^{2}\frac{b}{2}-2=$

$\frac{2}{1+{s}^{2}}+\frac{2}{1+{t}^{2}}-2=1$

RizerMix

Answered 2022-01-20
Author has **438** answers

alenahelenash

Answered 2022-01-24
Author has **343** answers

Hint: $\frac{1-{s}^{2}}{1+{s}^{2}}+\frac{1-{t}^{2}}{1+{t}^{2}}=1$
$\iff \frac{1-{s}^{2}}{1+{s}^{2}}=1-\frac{1-{t}^{2}}{1+{t}^{2}}$ $\iff (1-{s}^{2})(1+{t}^{2})=2{t}^{2}(1+{s}^{2})$

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Solve this equation for$-{180}^{\circ}\le \theta \le {180}^{\circ}$

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How to graph $f\left(x\right)=\frac{1-\mathrm{cos}(2x-2)}{{(x-1)}^{2}}$ analytically?

I only have an issue when it comes to determining the hole at $x=1$ analytically. Graphing the function on a graphing utility reveals a hole at $x=1$ rather than a vertical asymptote. I know that by simply plugging in the value $1$ for $x$, one will see that both the numerator and denominator equate to zero. This shows that there is at least one factor of $(x-1)$ that can be extracted from the numerator.

I would like to know how to simplify this function analytically in such a way that two factors of $(x-1)$ can be obtained from the numerator. I believe one must make use of an inverse function down the line in order to extract the obvious $(x-1)$ hiding in the $\mathrm{cos}(2x-2)=\mathrm{cos}\left(2(x-1)\right)$ term. That is as far as I got in the problem.

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Prove trigonometric equivalence

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Solving $\mathrm{sin}\left(\frac{x}{{x}^{2}+1}\right)+\mathrm{sin}\left(\frac{1}{{x}^{2}+x+2}\right)=0$

Here is my way:

First write the equation as

$\mathrm{sin}(\frac{x}{2({x}^{2}+1)}+\frac{1}{2({x}^{2}+x+2)})\mathrm{cos}(\frac{x}{2({x}^{2}+1)}-\frac{1}{2({x}^{2}+x+2)})=0$

Then the real solution can be find by

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which gives x=-1.

Here is my way:

First write the equation as

Then the real solution can be find by

which gives x=-1.

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How can I express y through R and x?

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It can be transformed into this :

$y=\pm R\mathrm{cos}\left(\frac{\sqrt{2yR-{y}^{2}}+x}{R}\right)-R$

How can I express y through R and x?

It can be transformed into this :