Trigonometry Problems - Basics, Advanced & Examples

A non-right triangle with known x,y coordinates. Lets say A,B,C is the points of triangle. I want to find a new point D which is perpendicular line from point B and lies on the line AC. Any simpled questions for D(x,y)?

trying to solve $\sqrt{\cos <!--\; \; -->(x)-<!--\; -\; -->2\cos <!--\; \; -->(2x)}+\sqrt{2}\cos <!--\; \; -->(2x)=0$
The system is
$\{\begin{array}{l}\cos <!--\; \; -->(x)-<!--\; -\; -->2\cos <!--\; \; -->(2x)=2{\cos }^2<!--\; \; -->(2x)\\ -<!--\; -\; -->\sqrt{2}\cos <!--\; \; -->(2x)\ge <!--\; \ge \; -->0⟺<!--\; ⟺\; -->\cos <!--\; \; -->(2x)\le <!--\; \le \; -->0\end{array}$
The equation:
$\cos <!--\; \; -->(x)-<!--\; -\; -->2(2{\cos }^2<!--\; \; -->(x)-<!--\; -\; -->1)=2(2{\cos }^2<!--\; \; -->(x)-<!--\; -\; -->1{)}^2$
$\cos <!--\; \; -->(x)-<!--\; -\; -->4{\cos }^2<!--\; \; -->(x)+2=2(4{\cos }^4<!--\; \; -->(x)-<!--\; -\; -->4{\cos }^2<!--\; \; -->(x)+1)$
$\cos <!--\; \; -->(x)-<!--\; -\; -->4{\cos }^2<!--\; \; -->(x)+2=8{\cos }^4<!--\; \; -->(x)-<!--\; -\; -->8{\cos }^2<!--\; \; -->(x)+2$
$8{\cos }^4<!--\; \; -->(x)-<!--\; -\; -->4{\cos }^2<!--\; \; -->(x)-<!--\; -\; -->\cos <!--\; \; -->(x)=0$
But what to do next?
If I factor out $\cos <!--\; \; -->(x)$, I get a non-factorizable third-degree polynomial in parentheses
$\cos <!--\; \; -->(x)(8{\cos }^3<!--\; \; -->(x)-<!--\; -\; -->4\cos <!--\; \; -->(x)-<!--\; -\; -->1)=0$

Given a non-right triangle with vertex $C={91}^{\circ <!--\; \circ \; -->},\stackrel{¯<!--\; ¯\; -->}{AC}=39.9cm,\stackrel{¯<!--\; ¯\; -->}{BC}=32.6cm$, find
a) $\stackrel{¯<!--\; ¯\; -->}{AB}$
b) $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}A$
c) $m\mathrm{\measuredangle <!--\; \measuredangle \; -->}B$
I was able to find the length of $\stackrel{¯<!--\; ¯\; -->}{AB}$, which I labeled $c$
$\begin{array}{rl}c=& \sqrt{{32.6}^2+{39.9}^2-<!--\; -\; -->2(32.6\ast <!--\; \ast \; -->39.9)\cos <!--\; \; -->({91}^{\circ <!--\; \circ \; -->})}\\ c=& 51.9cm\end{array}$
but I'm having a hard time finding the angle measures of $A$ and $B$, here's my attempt:
a) $\begin{array}{rl}\frac{32.6}{\sin <!--\; \; -->A}=& \frac{51.9}{\sin <!--\; \; -->{91}^{\circ <!--\; \circ \; -->}}\\ \\ \sin <!--\; \; -->A=& \frac{32.6\sin <!--\; \; -->{91}^{\circ <!--\; \circ \; -->}}{51.9}\end{array}$
So given some theta A, we get the vertical length 0.63 radians. To find A, I used
$\begin{array}{r}\arcsin <!--\; \; -->(0.63)\approx <!--\; \backslash approx\; -->0.68\text{rad}\end{array}$
So I figured I just needed to convert this to degree measure, getting me
$m\mathrm{\measuredangle <!--\; \measuredangle \; -->}A\approx <!--\; \backslash approx\; -->{38.96}^{\circ <!--\; \circ \; -->}$
b) using the fact that the sum of angle measures of a triangle is ${180}^{\circ <!--\; \circ \; -->}$, I tried to solve for angle $B$ using
$\begin{array}{rl}{38.96}^{\circ <!--\; \circ \; -->}+{51.9}^{\circ <!--\; \circ \; -->}+m\mathrm{\measuredangle <!--\; \measuredangle \; -->}B\approx <!--\; \backslash approx\; -->& {180}^{\circ <!--\; \circ \; -->}\\ m\mathrm{\measuredangle <!--\; \measuredangle \; -->}B\approx <!--\; \backslash approx\; -->& {89.14}^{\circ <!--\; \circ \; -->}\end{array}$
but I'm getting marked wrong for both answers b) and c). What am I doing wrong?

Prove the following:
$(\genfrac{}{}{0ex}{}{n}{0})-<!--\; -\; -->(\genfrac{}{}{0ex}{}{n}{2})+(\genfrac{}{}{0ex}{}{n}{4})-<!--\; -\; -->(\genfrac{}{}{0ex}{}{n}{6})+...=2^{\frac{n}{2}}\cos <!--\; \; -->\frac{n\mathrm{\pi <!--\; \pi \; -->}}{4}$

Is every triangle with 3 rational angles (in degrees) similar to one with 3 integer-length sides?

I needed to solve the following equation:
$\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}+\tan <!--\; \; -->3\mathrm{\theta <!--\; \theta \; -->}=\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->3\mathrm{\theta <!--\; \theta \; -->}$
Transform the LHS first:
$\begin{array}{rl}\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}+\tan <!--\; \; -->3\mathrm{\theta <!--\; \theta \; -->}& =(\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->})+{\displaystyle \frac{\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}}{1-<!--\; -\; -->\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}}}\\ & ={\displaystyle \frac{(\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->})(2-<!--\; -\; -->\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->})}{1-<!--\; -\; -->\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}}}\end{array}$
And, RHS yields
$\begin{array}{rl}\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->3\mathrm{\theta <!--\; \theta \; -->}& =(\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}){\displaystyle \frac{\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}}{1-<!--\; -\; -->\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}}}\end{array}$
Now, two terms can be cancelled out from LHS and RHS, yielding the equation:
$\begin{array}{rl}2-<!--\; -\; -->\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}& =\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}\\ \tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\tan <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}& =1,\end{array}$
which can be further reduced as:
${\tan }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\frac{1}{3}⟹<!--\; ⟹\; -->\tan <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=\pm <!--\; \pm \; -->\frac{1}{\sqrt{3}}$
Now, we can yield the general solution of this equation:
$\mathrm{\theta <!--\; \theta \; -->}=n\mathrm{\pi <!--\; \pi \; -->}\pm <!--\; \pm \; -->{\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{6}},n\in <!--\; \in \; -->Z$. But, setting $\mathrm{\theta <!--\; \theta \; -->}={\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{6}}$ in the original equation is giving one term $\tan <!--\; \; -->{\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}$, which is not defined.

Is there a way to prove this relation?:
$\underset{\text{n}=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{{\text{s}}^2+{(1+2\text{n})}^2{\mathrm{\omega <!--\; \omega \; -->}}^2}=\frac{\mathrm{\pi <!--\; \pi \; -->}\tanh <!--\; \; -->\left(\frac{\mathrm{\pi <!--\; \pi \; -->}\text{s}}{2\mathrm{\omega <!--\; \omega \; -->}}\right)}{4\text{s}\mathrm{\omega <!--\; \omega \; -->}}$
And find the conditions for which this equality hold?

I am trying to show that for every $\mathrm{\lambda <!--\; \lambda \; -->}\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->},\mathrm{\pi <!--\; \pi \; -->})\setminus <!--\; \setminus \; -->\{0\}$ and every $n\in <!--\; \in \; -->{\mathbb{Z}}_{>0}$
$\frac{1}{n}\sqrt{\frac{1-<!--\; -\; -->\cos <!--\; \; -->(n-<!--\; -\; -->1)\mathrm{\lambda <!--\; \lambda \; -->}}{1-<!--\; -\; -->\cos <!--\; \; -->\mathrm{\lambda <!--\; \lambda \; -->}}}\le <!--\; \le \; -->1$

How would I solve the following trigonometric equation?
$2\cot <!--\; \; -->2x\cos <!--\; \; -->2x=1-<!--\; -\; -->\sin <!--\; \; -->2x$
I got to this stage: $2{\cos }^2<!--\; \; -->2x=\sin <!--\; \; -->2x-<!--\; -\; -->{\sin }^2<!--\; \; -->2x$
How do I continue?

Let $\mathrm{\alpha <!--\; \alpha \; -->}$, $\mathrm{\beta <!--\; \beta \; -->}$ and $\mathrm{\gamma <!--\; \gamma \; -->}$ be the three angles in a non-right triangle. How can i prove the following inequality?
${\displaystyle \frac{1+{\sin }^2<!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})}{{\cos }^2<!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})}}+{\displaystyle \frac{1+{\sin }^2<!--\; \; -->(\mathrm{\beta <!--\; \beta \; -->})}{{\cos }^2<!--\; \; -->(\mathrm{\beta <!--\; \beta \; -->})}}+{\displaystyle \frac{1+{\sin }^2<!--\; \; -->(\mathrm{\gamma <!--\; \gamma \; -->})}{{\cos }^2<!--\; \; -->(\mathrm{\gamma <!--\; \gamma \; -->})}}\ge <!--\; \ge \; -->{\displaystyle \frac{1+\sin <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})\sin <!--\; \; -->(\mathrm{\beta <!--\; \beta \; -->})}{1-<!--\; -\; -->\sin <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})\sin <!--\; \; -->(\mathrm{\beta <!--\; \beta \; -->})}}+{\displaystyle \frac{1+\sin <!--\; \; -->(\mathrm{\beta <!--\; \beta \; -->})\sin <!--\; \; -->(\mathrm{\gamma <!--\; \gamma \; -->})}{1-<!--\; -\; -->\sin <!--\; \; -->(\mathrm{\beta <!--\; \beta \; -->})\sin <!--\; \; -->(\mathrm{\gamma <!--\; \gamma \; -->})}}+{\displaystyle \frac{1+\sin <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})\sin <!--\; \; -->(\mathrm{\gamma <!--\; \gamma \; -->})}{1-<!--\; -\; -->\sin <!--\; \; -->(\mathrm{\alpha <!--\; \alpha \; -->})\sin <!--\; \; -->(\mathrm{\gamma <!--\; \gamma \; -->})}}$

Prove that
$3x-<!--\; -\; -->x^3<\frac{2}{\sin <!--\; \; -->2x},\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->(0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$
I have tried by proving that
$3x-<!--\; -\; -->x^3<\frac{9}{5\mathrm{\pi <!--\; \pi \; -->}}x+\frac{3}{2}<\frac{2}{\sin <!--\; \; -->2x},\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->(0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$
with Jensen's inequality, but I hoped this problem would have a simpler solution.

$A=(x_1,y_1)$, $B=(x_2,y_2)$, and $C=(x_3,y_3)$ are three (distinct) non-collinear points in the Cartesian plane, and $a=\left|\stackrel{¯<!--\; ¯\; -->}{BC}\right|$, $b=\left|\stackrel{¯<!--\; ¯\; -->}{AC}\right|$, and $c=\left|\stackrel{¯<!--\; ¯\; -->}{AB}\right|$. The incenter of the triangle is
$(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}).$
The x-coordinate of the incenter is a "weighted average" of the x-coordinates of the vertices of the given triangle, and the y-coordinate of the incenter is the same "weighted average" of the y-coordinates of the same vertices. I am requesting an explanation for this statement.

Given a triangle $ABC$, $\mathrm{\angle <!--\; \angle \; -->}C$ is $65$ degrees, side $C$ is 10. The area of the triangle is $20.$ What is the perimeter?

Trigonometry equation $\cos <!--\; \; -->(x)-<!--\; -\; -->\cos <!--\; \; -->(2x)+\cos <!--\; \; -->(3x)=0$

I am given 2 side lengths for one triangle and two side lengths for the parallelogram. I am asked to find the length of m (FE) and n (DE)
given the lenghts:
h (AC) = 9
k (AF) = 15
f (AB) = 16
I don't see how to use Law of Sines because I don't have any angles and I don't see how to use Law of Cosines to solve triangle ACF because I am missing a side length.

Consider a non-isosceles triangle, pick a vertex. Assume that the median and the altitude passing through this vertex are isogonal conjugates (i.e. symmetric w.r.t. the bisector of the angle). Prove that the triangle is right-angled in this vertex.

Evaluate ${\csc }^{-<!--\; -\; -->1}<!--\; \; -->(\sec <!--\; \; -->5)$ without using a calculator.

How to solve goniometric equation where function arguments differ?
${\cos }^2<!--\; \; -->x+\cos <!--\; \; -->2x+1=0$

$\sin <!--\; \; -->x+\sin <!--\; \; -->2x+\sin <!--\; \; -->3x=1+\cos <!--\; \; -->x+\cos <!--\; \; -->2x$ - solving a trigonometric equation