P(x, y) is the endpoint of the unit circle determined by t. Then $\mathrm{sin}t=,\mathrm{cos}t=,\text{}\text{and}\text{}\mathrm{tan}t=$.

a2linetagadaW
2021-08-20
Answered

P(x, y) is the endpoint of the unit circle determined by t. Then $\mathrm{sin}t=,\mathrm{cos}t=,\text{}\text{and}\text{}\mathrm{tan}t=$.

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joshyoung05M

Answered 2021-08-21
Author has **97** answers

According to the definition, $\mathrm{sin}t$ is equivalent to y. $\mathrm{cos}t$ is equivalent to x. $\mathrm{tan}t$ is equivalent to $\frac{y}{x}$.

The unidentified terms are $x,y,\frac{y}{x}$.

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How does $\frac{-{e}^{-x}}{\sqrt{{e}^{-2x}-1}}$ rearrange?

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Injectivity and range of $\mathrm{arctan}\left(\sqrt{\frac{1+x}{1-x}}\right)$

According to the definition, for a function to be injective$f\left(a\right)=f\left(b\right)\to a=b$ for all $a,b\in {D}_{f}$

Using this I get:

$\mathrm{arctan}\left(\sqrt{\frac{1+a}{1-a}}\right)=\mathrm{arctan}\left(\sqrt{\frac{1+b}{1-b}}\right)$

$\iff \left(\sqrt{\frac{1+a}{1-a}}\right)=\left(\sqrt{\frac{1+b}{1-b}}\right)$

$\iff \frac{1+a}{1-a}=\frac{1+b}{1-b}$

$\iff 2a=2b$

$a=b$

Hence the function is injective. Now this could be total bs but I also am not able to find the proper range for this one. WolframAlpha says$0\le y<\frac{\pi}{2}$ . But how does one come up with this range?

According to the definition, for a function to be injective

Using this I get:

Hence the function is injective. Now this could be total bs but I also am not able to find the proper range for this one. WolframAlpha says

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Prove that:

$\frac{{\mathrm{sin}}^{2}\left(2\alpha \right)}{{\mathrm{sin}}^{2}\left(\alpha \right)}=4-4{\mathrm{sin}}^{2}\left(\alpha \right)$

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Simplification of ${\mathrm{cos}}^{4}\left(x\right)+{\mathrm{sin}}^{4}\left(x\right)$

$\left(\mathrm{sin}x\right)}^{4}+{\left(\mathrm{cos}x\right)}^{4}=\frac{{(1-\mathrm{cos}2x)}^{2}}{4}+\frac{{(1+\mathrm{cos}2x)}^{2}}{4$

$=\frac{1-2\mathrm{cos}2x+{\left(\mathrm{cos}2x\right)}^{2}+1+2\mathrm{cos}2x+{\left(\mathrm{cos}2x\right)}^{2}}{4}$

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

its correct?

its correct?

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I wanted to use this

$\mathrm{sin}\left(\frac{\pi}{15}\right)\cdot \mathrm{sin}\left(2\frac{\pi}{15}\right)\dots \mathrm{sin}\left(7\frac{\pi}{15}\right)=\sqrt{15}$

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Why does $\mathrm{tan}(\frac{\pi}{2}-\beta )=\frac{t}{s}$ imply that $\mathrm{tan}\beta =\frac{s}{t}$?

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When working with spherical coordinates I have come across

${\mathrm{cos}}^{-1}(\mathrm{cos}\left(a\right)\times \mathrm{cos}\left(b\right))$

and was wondering if there was some kind of identity for this that I could use instead of computing the inside of the acrccos function first.

and was wondering if there was some kind of identity for this that I could use instead of computing the inside of the acrccos function first.