Solving $2\mathrm{sin}(x+{30}^{\circ})=\mathrm{cos}(x+{150}^{\circ})$ for x between $0}^{\circ}\text{}\text{and}\text{}{360}^{\circ$

joygielymmeloiy
2022-01-24
Answered

Solving $2\mathrm{sin}(x+{30}^{\circ})=\mathrm{cos}(x+{150}^{\circ})$ for x between $0}^{\circ}\text{}\text{and}\text{}{360}^{\circ$

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spelkw

Answered 2022-01-25
Author has **12** answers

HINT

The equation is equivalent to

$2\mathrm{sin}x{\mathrm{cos}30}^{\circ}+2{\mathrm{sin}30}^{\circ}\mathrm{cos}x=\mathrm{cos}x{\mathrm{cos}150}^{\circ}-\mathrm{sin}x{\mathrm{sin}150}^{\circ}$

$\sqrt{3}\mathrm{sin}x+\mathrm{cos}x=-\frac{\sqrt{3}}{2}\mathrm{cos}x-\frac{12}{\mathrm{sin}x}$

The equation is equivalent to

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Let P(x, y) be the terminal point on the unit circle determined by t. Then

asked 2022-01-25

In any triangle is $\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C=\frac{3\sqrt{3}}{2}$ always
Well, I came with an interesting proof. But I just want to verify it
Applied at function $y=\mathrm{sin}x$
We have, $\mathrm{sin}(\frac{A+B+C}{2})\ge \frac{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}{3}$
From here we will get
$\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\le \frac{3\sqrt{3}}{2}$
Also by A.M. $\ge $ G.M. in an acute angled triangle
$\frac{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}{3}\ge \sqrt[3]{\mathrm{sin}A\mathrm{sin}B\mathrm{sin}C}$
$\Rightarrow \mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\ge 3(\sqrt[3]{\mathrm{sin}A\mathrm{sin}B\mathrm{sin}C})$
$\Rightarrow \mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\ge 3(\frac{\sqrt{3}}{2})=\frac{3\sqrt{3}}{2}>2$
and from this I get
$\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\ge \frac{3\sqrt{3}}{2}$

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I have an impulse train given by

$\frac{1}{R+1}+\frac{\sum _{k=1}^{R}\mathrm{cos}\left(\frac{2k\pi x}{R+1}\right)}{R+1}$

It seems obvious to me that, for x=0, the function returns 1. This is because$\mathrm{cos}\left(0\right)=1$ , and we therefore end up with $\frac{1}{R+1}+\frac{R}{R+1}=\frac{R+1}{R+1}=1$

However indeterminate result at x=0. Usually this means there is a division by 0 somewhere. But I can't see any reason for this function to produce an indeterminate result.

It seems obvious to me that, for x=0, the function returns 1. This is because

However indeterminate result at x=0. Usually this means there is a division by 0 somewhere. But I can't see any reason for this function to produce an indeterminate result.

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Using De Moivre's formula:

${(\mathrm{cos}\theta +i\mathrm{sin}\theta )}^{4}=\mathrm{cos}4\theta +i\mathrm{sin}4\theta$

Using De Moivre's formula:

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