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Don S.
Teacher / Tutor for 12 years
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Geology
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Question:

Which mineral (often found in a box in an undergraduate lab) sticks to your tongue if you lick it?

Don S.

Kaolinite.

Pre-Calculus
TutorMe
Question:

Simplify $$(\log_8 27)(\log_9 16)$$.

Don S.

We proceed by using a change of base of logarithms. Recall $$\log_y x = \frac{\log_b x}{\log_b y}$$. Thus, our problem simplifies as follows $$(\log_8 27)(\log_9 16) = \frac{\log27\, \cdot\, \log16}{\log8 \, \cdot \, \log9}$$ $$= \frac{\log3^3\, \cdot\, \log2^4}{\log2^3\, \cdot\, \log3^2}$$ $$=\frac{3\log3\, \cdot\, 4\log2}{3\log2\, \cdot\, 2\log3}$$ $$= 2$$

Trigonometry
TutorMe
Question:

Find an algebraic expression for the somewhat intimidating-looking function $$f(x)=\sin(\tan^{-1}x)$$. Note that you are being asked to write down this expression in terms of only $$x$$, and not using any trigonometric functions.

Don S.

For some reason, people tend to hate these kinds of questions. I like them; I am odd--I know this. In any case, let's look at this problem in a slightly simpler form. It looks very much like $$\sin(\theta)$$ if we let $$\theta=\tan^{-1}x$$. Enjoying this new name for the argument of the sine function, we see immediately, by taking the tangent of both sides of the equation, that $$x=\tan\theta$$. This is where a picture is truly worth a thousand words, but they aren't seeming to let me draw one in this space :( If it were there, it would be a simple right triangle drawn in the first quadrant, with angle $$\theta$$ in the lower left corner. This is all we need! Since $$\tan\,\theta=x$$ this triangle has vertical side length $$x$$, and horizontal side length 1 (since $$\tan\,\theta = \text{opposite / adjacent}$$). By the Pythagorean Theorem, the hypotenuse of the triangle must be $$\sqrt{1+x^2}$$. We are nearly finished. The original question requested an algebraic expression for $$\sin(\tan^{-1}x)$$, we called this $$\sin\,\theta$$ and by our picture, we see that $$f(x) = \sin(\tan^{-1}x) = \sin\,\theta =\frac{x}{\sqrt{1+x^2}}$$ (since $$\sin\,\theta = \text{opposite / hypotenuse}$$). Now, wasn't that fun?! (Next time we will do this with a double angle identity for extra fun...)

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