# Tag Info

### length of hypotenuse where only part know

I would say the simplest way to work this out is but using scale factors. You can divide the section $L=35+x$, and now a scale factor is $15/8 = 1.875$. Thus, $(35+x)/1.875 = x$. Since scale factor is ...

### Showing $\frac{\frac1{\cos A}-\frac1{\sin A}}{\frac1{\cos A}+\frac1{\sin A}}=\frac{\frac{\sin A}{\cos A}-1}{\frac{\sin A}{\cos A}+1}$

@Ned answered this question as simply as possible; "Multiply top and bottom by sin A" I can't believe I didn't see that. Thank you Ned.
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### Find number of solutions to the equation $\sin(6\sin x)=\frac{x}{6}$.

If $x$ is a root of the equation, so is $-x$. Furthermore, $x=0$ is a root. For these reasons, we only enumerate the number of positive roots. Since $\frac{x}{6}>1$ for $x>6$, for every positive ...
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Accepted

### What am I doing wrong, when calculating sequence limit $\left|\sin \left(\pi \cdot \sqrt[3]{n^3+n^2}\right)\right|$

For $b_n$ which satisfies $b_n \to 1$, $a_n$ be $$a_n = \left|\sin(\pi n b_n)\right|$$ Your attempt is Since $b_n \to 1$, $\pi nb_n$ may be approximated to $$\pi nb_n \approx \pi n$$ which is False. ...
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### Question regarding cuberoots of integral multiples of i

If you write $z^{3}=8i$ as $z^{3}=(-2i)^{3}$ (since $i^3=-i$) you will see that the scale is by factor 2. So the point you are looking at is $(2\cos{\frac{\pi}{6}},2\sin{\frac{\pi}{6}})$ and the ...
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### Find value of $2A$ if $A=\frac{3\tan\left(A\right)}{1-\tan\left(A\right)}-1$

Let us start by inverting the homographic relationship : $$A=\frac{4 \tan A -1}{1- \tan A} \ \iff \ \tan A=\frac{A+1}{A+4}=1-\frac{3}{A+4}$$ The idea is to use the following expansion of the "...
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### How to get the size of the angle bisector of a triangle using the law of cosines

Thanks for showing your work. The angle bisector theorem gives us $AC \times BD = AB \times CD$ like you mentioned. Notice that your cosine rule equations can be written as (EG Clearing denominators ...
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### Does $\frac{\sin(\theta-\alpha)}{\sin\alpha}=\frac{\cos(\theta+\gamma-\alpha)}{\cos(\gamma-\alpha)}$ have an analytical solution for $\alpha$?

Short answer: $$\alpha=\arctan\left(\frac1{\cot\theta-\tan\gamma+\sqrt{\tan^2\gamma+\cot^2\theta+1}}\right)$$ is the positive $\alpha.$ Directly from the geometric question, if we write in coordinate ...
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### Is it true that $\sin^2((2n+1) \pi x) \geq c$ occurs often (in precise sense)?

The claim is true. Indeed, consider the sequence $(z_n)_{n\in\mathbb N}$ of points of the unit circle $$\mathbb T=\{(u,v)\in\mathbb R^2:u^2+v^2=1\}$$ such that $z_n=(\cos (2n+1)\pi x,\sin (2n+1)\pi x)$...
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### General solution of $\cos(3\theta) - \cos(\theta) = 0$

Plugging in the candidate solution angles into the original equation, we see that any integer multiple of $\frac{\pi}{2}$ (90°) will satisfy it. Your two solutions are actually the same. Method 1 ...
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### General solution of $\cos(3\theta) - \cos(\theta) = 0$
You first solution set $\{\frac{(2n+1)\pi}{2},n\pi\}$ is equal to the second $\{\frac{n\pi}{2},n\pi\}$.