# Tag Info

Accepted

### Simplifying $\frac{1-\cos^2(\alpha)}{\sin(\alpha) \cos(\alpha)}$ until there is only one trigonometric function left

You can use the fact that \begin{align} \frac{1-\cos^2(\alpha)}{\sin(\alpha)\cos(\alpha)}&=\frac{\sin^2(\alpha)}{\sin(\alpha)\cos(\alpha)}\\ &=\frac{\sin(\alpha)}{\cos(\alpha)}\\ &=\tan(\...
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### Simplifying $\frac{1-\cos^2(\alpha)}{\sin(\alpha) \cos(\alpha)}$ until there is only one trigonometric function left

You must know the Pythagorean trigonometric identity $$\sin^2(\alpha)+\cos^2(\alpha)=1\implies 1-\cos^2(\alpha)=\sin^2(\alpha).$$ From there, use the definition of $\tan(\alpha)$ that you already ...
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### Question inside $\int \tan^2 x dx$

This is a phenomenon that happens with all indefinite integrals. The original integral $$\int \tan^2 x\:dx$$ has singularities at $\frac{\pi}{2} + \pi k$ for $k\in\Bbb{Z}$, therefore, no, the original ...
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### In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$

Here's my approach: 1.) First, we draw a line $AD$ from $A$ such that $\angle ABD=\angle DAB=\beta$, this means that $\angle DAC=\alpha-\beta$ ($\angle BAC=\alpha$). This implies that $AD=BD=x$. Now, ...
• 2,102
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### Why does cancelling by $\sin x$ when solving $4\tan x = 5\sin x$ for $0\leq x < 2\pi$ miss solutions?

For the same reason why solving $x = x^2$ by dividing by $x$ would lose solutions. If you divide both sides by something, you implicitly give rise to two cases. For $x=x^2$, if you divide by $x$, then ...
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### Determine relative position of 3 large (equal) circles and 1 smaller circle within a minimum enclosing circle

For Case 1, using Descartes' Kissing Circles Theorem, we can determine the maximum possible value of $r_2$. From the $r_1$ circles forming an equilateral triangle, it follows that the enclosing ...

### Proving that, if $\cos2\alpha$ is irrational, then $\sin\alpha$ and $\tan\alpha$ are irrational

I am not sure why the previous answer got downvoted, but contradiction is the easiest way forward. You can also use contraposition, the arguments are pretty much the same. Start by assuming ...
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### Is this u-substitution valid?

Of course squaring still yields a true identity, but as your comment regarding absolute value suggests (and as Brian Moehring made explicit in the comments) $\cos x = \sqrt{1 - u^2}$ cannot hold on ...
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1 vote

### Why does cancelling by $\sin x$ when solving $4\tan x = 5\sin x$ for $0\leq x < 2\pi$ miss solutions?

We can’t cancel terms on both sides without knowing whether it is zero. Just like solving $x^2=x \Leftrightarrow x^2-x=0 \Leftrightarrow x(x-1)=0 \Leftrightarrow x=0 \textrm{ or }x=1, \tag*{}$ we ...
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### Reduce a linear combination of sin and cos to a single trigonometric function

Hint You want to express $a\sin x+b\cos x$ in terms of a single sine (or cosine). Since the periodicity does not change, let $a\sin x+b\cos x = A\sin (x+k)$ with $A,k$ to be determined.

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