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15
Math.TechQA.Club
2026-03-25 10:15:45
57
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Geometry Question (Tips)
Published on
25 Mar 2026 - 10:15
#geometry
#trigonometry
#euclidean-geometry
#triangles
93
Views
Basic trigonometry - $\sin{(-a)} = -\sin{(a)}$
Published on
01 Mar 2018 - 22:50
#trigonometry
110
Views
Evaluating $\int_{-\infty}^\infty \sin(x)\cos(x)\ dx$
Published on
26 Mar 2026 - 1:00
#integration
#trigonometry
#improper-integrals
52
Views
How to calculate the total lenght of a series of arc lenghts when the angle changes?
Published on
02 Mar 2018 - 2:21
#algebra-precalculus
#trigonometry
136
Views
What is the maximum value of $k$ so that $x^2+4(\sin^{2}{x}\tan^{2}{x}+\cos^{2}{x}\cot^{2}{x}+k^2-x\sec{x}\csc{x})=0$ has real roots.
Published on
28 Mar 2026 - 12:14
#trigonometry
#polynomials
155
Views
Prove $\sin8\theta-\sin10\theta=\cot9\theta(\cos10\theta-\cos8\theta)$
Published on
02 Mar 2018 - 3:27
#algebra-precalculus
#trigonometry
292
Views
If $a,b,c$ are in arithmetic progression, prove that...
Published on
26 Mar 2026 - 10:44
#sequences-and-series
#trigonometry
#fractions
#arithmetic-progressions
2.1k
Views
number of distinct solution $x\in[0,\pi]$ of the equation satisfy $8\cos x\cos 4x\cos 5x=1$
Published on
25 Mar 2026 - 1:18
#trigonometry
#polynomials
#roots
#factoring
#substitution
99
Views
Tower on a slope problem
Published on
02 Mar 2018 - 9:53
#trigonometry
3k
Views
How to visualise positive and negative tangents
Published on
01 Apr 2026 - 12:01
#trigonometry
#visualization
#tangent-line
69
Views
how to solve this trig equation?
Published on
02 Mar 2018 - 15:52
#trigonometry
141
Views
Finding the exact value of $\tan\left(\frac{1}{2}\arccos \frac{2}{5}\right)$
Published on
02 Mar 2018 - 19:53
#trigonometry
68
Views
Adjusting the radii of two circles, with known distance between circles
Published on
25 Mar 2026 - 14:14
#geometry
#trigonometry
#circles
#average
235
Views
Integration problem involving $\sqrt{\cos(x)-\cos^3(x)}$
Published on
24 Mar 2026 - 23:44
#real-analysis
#trigonometry
#definite-integrals
#radicals
#substitution
84
Views
$e^{-\beta \frac{1}{2}\hbar \omega} \frac{1}{1 - e^{\beta \hbar \omega}} = \frac{1}{2 \sinh \left( \frac{\beta \hbar \omega}{2}\right)}$
Published on
25 Mar 2026 - 3:03
#trigonometry
#hyperbolic-functions
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