The mathematician formerly known as DayLateDon

19h
revised Find the area of a triangle given the radius of its incircle and a tangential point
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20h
answered Find the area of a triangle given the radius of its incircle and a tangential point
22h
revised Does the centroid of a triangle ever fall outside of its Morley's triangle?
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23h
answered Does the centroid of a triangle ever fall outside of its Morley's triangle?
1d
comment Calculation of integral using two different methods?
This is a subtle point, but an important one ... In verifying that the difference between the two solutions is a constant, you should not use the same "$+C$" for both instances and allow them to cancel; they aren't necessarily "the same constant". Instead, you should argue in more generality: $$(\text{this} + C_1) - (\text{that} + C_2) = \text{this} - \text{that} + C_1 - C_2 = -\frac{1}{4} + C_1 - C_2$$ Since $C_1$ and $C_2$ are constants, the final expression is one, too.
1d
comment Calculation of integral using two different methods?
To complete the discussion and show that the two solutions are consistent with each other, subtract one from the other and note that the difference is a constant.
1d
answered Trying to solve a pair of trigonometric simultaneous equations
1d
comment Sum/Difference Identity Formula Question
You have the right idea. However, the equation has $13\pi/15$, whereas your solution uses $13\pi/5$ (that is, the denominators are different). Based on the equation as written, the solution should be $$\cos\left(\frac{13\pi}{15}+\left(-\frac{\pi}{5}\right)\right) = \cos\left(\frac{13\pi}{15}-\frac{3\pi}{15}\right) = \cos\frac{10\pi}{15}= \cos\frac{2\pi}{3}$$
2d
comment Prove that a trigonometric equation has six distinct roots
Descartes' Rule of Signs also indicates problematic cases. For instance, if $C+B$, $C-B$, and $A$ are all positive, then there are either $2$ negative roots, or none at all. Or, if $A=0$, while $C+B$ and $C-B$ match in sign (but aren't themselves zero), then there are no real roots. And so forth. The possibility of six roots requires that $B+C$ and $B-C$ have different signs.
2d
comment Given 4 points with 2 on different radius. Obtain the center of the circle.
What geometric relationship exists between the black hole and two positions of a particular star? (In other words: How does the center of a circle relate to two points on that circle?)
2d
revised Finding a triangle ABC if $2\prod (\cos \angle A+1)=\sum \cos(\angle A-\angle B)+\sum \cos \angle A+2$
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2d
revised Finding a triangle ABC if $2\prod (\cos \angle A+1)=\sum \cos(\angle A-\angle B)+\sum \cos \angle A+2$
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2d
answered For two vectors $a$ and $b$, why does $\cos(θ)$ equal the dot product of $a$ and $b$ divided by the product of the vectors' magnitudes?
2d
comment Finding a triangle ABC if $2\prod (\cos \angle A+1)=\sum \cos(\angle A-\angle B)+\sum \cos \angle A+2$
@sepideh: Actually, there's another solution.
2d
answered Finding a triangle ABC if $2\prod (\cos \angle A+1)=\sum \cos(\angle A-\angle B)+\sum \cos \angle A+2$
2d
comment Finding a triangle ABC if $2\prod (\cos \angle A+1)=\sum \cos(\angle A-\angle B)+\sum \cos \angle A+2$
Have you tried anything?
Jul
29
revised Trig identity involving sum of cosines
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Jul
29
comment What is the equation describing a three dimensional, 14 point Star?
@ThePolywellGuy: "The actual solution is:" ... well?
Jul
29
awarded Yearling
Jul
29
awarded Yearling
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