Trigonometry Problems - Basics, Advanced & Examples

The height of a vertical tower is to be found by a surveyor. The angle of elevation of the top of the tower from a point on the horizontal ground some distance away is measured as 28.7 degrees. From this point the surveyor moves 21.6 metres directly towards the tower and repeats the measurement. The angle of elevation of the top of the tower from this point is 68.4 degrees. Find the height of the tower.

$\sin <!--\; \; -->x\cdot <!--\; \cdot \; -->\sin <!--\; \; -->2x\cdot <!--\; \cdot \; -->\sin <!--\; \; -->3x=\frac{\sin <!--\; \; -->4x}{4}$ - solving a trigonometric equation

How many solutions exist for the equation $2\sin <!--\; \; -->(x)+\cos <!--\; \; -->(x)=\sqrt{3}$ in $[0,2\mathrm{\pi <!--\; \pi \; -->}]$?

Establish identitiy $(\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2+(\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\sin <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2=2$

Solve for $\mathrm{\theta <!--\; \theta \; -->}$ $[0°<\mathrm{\theta <!--\; \theta \; -->}<180°]$
$\sin <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}+\sin <!--\; \; -->4\mathrm{\theta <!--\; \theta \; -->}=\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+\cos <!--\; \; -->3\mathrm{\theta <!--\; \theta \; -->}.$

Would a right triangle with bases a=i and b=1 have hypotenuse c=0?

Solve the trignometric equation $2\sin <!--\; \; -->(3x+\mathrm{\pi <!--\; \pi \; -->}/4)=\sqrt{1+8\sin <!--\; \; -->2x{\cos }^2<!--\; \; -->x}$

Solving trigonometric equation $\sin <!--\; \; -->(x)+\tan <!--\; \; -->(x)=0$

The hyperbolic sine function, $\sinh <!--\; \; -->(x)$ , is defined by the equation:
$\sinh <!--\; \; -->(x)=\frac{e^x-<!--\; -\; -->e^{-<!--\; -\; -->x}}{2}$
Find a formula for its inverse,
${\sinh }^{-<!--\; -\; -->1}<!--\; \; -->(x)$

Using expansion of $\sin <!--\; \; -->8\mathrm{\theta <!--\; \theta \; -->}$ to find equation with given roots

Find $\mathrm{\theta <!--\; \theta \; -->}$ from the equation $\cos <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}\cdot <!--\; \cdot \; -->\cos <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}\cdot <!--\; \cdot \; -->\cos <!--\; \; -->3\mathrm{\theta <!--\; \theta \; -->}=1/4$

By doing some right triangle gymnastics, we can derive things like
$\cos <!--\; \; -->(\arctan <!--\; \; -->x)=\frac{1}{\sqrt{1+x^2}}$, for $x>0$ $\cos <!--\; \; -->(\arcsin <!--\; \; -->x)=\sqrt{1-<!--\; -\; -->x^2}$
$\cos <!--\; \; -->(\arcsin <!--\; \; -->x)=\sqrt{1-<!--\; -\; -->x^2}$
$\tan <!--\; \; -->(\arcsin <!--\; \; -->x)=\frac{x}{\sqrt{1-<!--\; -\; -->x^2}}$
What about $\arctan <!--\; \; -->\cos <!--\; \; -->(x)$, $\arcsin <!--\; \; -->(\tan <!--\; \; -->x)$, etc?

Proving the identity $\frac{{\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}+{\tan }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->1}{{\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}={\tan }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}$

Let $f(x)=p\cos <!--\; \; -->x+q\sin <!--\; \; -->x,|p|+|q|\ne <!--\; \ne \; -->0$ where $p,q\in <!--\; \in \; -->R$ and $|f(x)|\le <!--\; \le \; -->1$.Let $\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}$ be the roots of the equation $f(x)=1,|\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}|=k\mathrm{\pi <!--\; \pi \; -->},k\in <!--\; \in \; -->R,$ then the find the possible values of k.

Verify this identity: $\sin <!--\; \; -->6\mathrm{\alpha <!--\; \alpha \; -->}+\sin <!--\; \; -->2\mathrm{\alpha <!--\; \alpha \; -->}=\frac{{\cos }^2<!--\; \; -->2\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->\cos <!--\; \; -->2\mathrm{\alpha <!--\; \alpha \; -->}\cos <!--\; \; -->6\mathrm{\alpha <!--\; \alpha \; -->}}{\sin <!--\; \; -->2\mathrm{\alpha <!--\; \alpha \; -->}}$

Solve trigonometric inequality $\sin <!--\; \; -->x+2\cos <!--\; \; -->x<2$
${\displaystyle \frac{2t}{1+t^2}}+2{\displaystyle \frac{1-<!--\; -\; -->t^2}{1+t^2}}<2$
$4t^2-<!--\; -\; -->2t>0$
$2t(2t-<!--\; -\; -->1)>0$
$t(2t-<!--\; -\; -->1)>0$
$(t>0\wedge <!--\; \wedge \; -->t>{\displaystyle \frac{1}{2}})\vee <!--\; \vee \; -->(t<0\wedge <!--\; \wedge \; -->t<{\displaystyle \frac{1}{2}})$
From this, I can only find $x<2\mathrm{\pi <!--\; \pi \; -->}+2k\mathrm{\pi <!--\; \pi \; -->}$, and, $x<2k\mathrm{\pi <!--\; \pi \; -->}$, these are good (I think), but I should find another two solutions.

Solve this equation :
$x^2+2x\cdot <!--\; \cdot \; -->\cos <!--\; \; -->b\cdot <!--\; \cdot \; -->\cos <!--\; \; -->c+{\cos }^2<!--\; \; -->b+{\cos }^2<!--\; \; -->c-<!--\; -\; -->1=0$
Such that $a+b+c=\mathrm{\pi <!--\; \pi \; -->}$ I don't have any idea. I can't try anything.

The value of ${\tan }^{-<!--\; -\; -->1}<!--\; \; -->\left(\frac{1}{\sqrt{2}}\right)-<!--\; -\; -->{\tan }^{-<!--\; -\; -->1}<!--\; \; -->\left(\frac{\sqrt{5-<!--\; -\; -->2\sqrt{6}}}{1+\sqrt{6}}\right)$
is equal to
$\frac{\mathrm{\pi <!--\; \pi \; -->}}{6}$
$\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$
$\frac{\mathrm{\pi <!--\; \pi \; -->}}{3}$
$\frac{\mathrm{\pi <!--\; \pi \; -->}}{12}$

What are some non-trivial ways to construct a similar triangle from a right triangle?
By non-trivial, I mean not:
- The triangle itself or any translations or rotations of it
- A 'shrinking' of the right triangle; that is, for $\mathrm{△<!--\; △\; -->}ABC$ with $\mathrm{\angle <!--\; \angle \; -->}ABC={90}^{\circ <!--\; \circ \; -->}$, given points $D$ on $\stackrel{¯<!--\; ¯\; -->}{AC}$ and $E$ on $\stackrel{¯<!--\; ¯\; -->}{BC}$ such that $\mathrm{\angle <!--\; \angle \; -->}DEC={90}^{\circ <!--\; \circ \; -->}$ we have $\mathrm{△<!--\; △\; -->}DEC\sim <!--\; \backslash sim\; -->\mathrm{△<!--\; △\; -->}ABC$, along with any arbitrary renaming of points. Or, more generally, any scaling of the triangle.
One of the most well-known ways to construct a similar triangle is by dropping a cevian from the hypotenuse to the vertex which is perpendicular to the hypotenuse. In other words, an altitude.
I can only think of one more case, and I'm not sure if it is true (verification would be appreciated); given right triangle $\mathrm{△<!--\; △\; -->}ABC$, let $D$ be the midpoint of $\stackrel{¯<!--\; ¯\; -->}{AC}$ and $E$ be on $\stackrel{¯<!--\; ¯\; -->}{BC}$ such that $\mathrm{\angle <!--\; \angle \; -->}BED={90}^{\circ <!--\; \circ \; -->}$. Then, $\mathrm{△<!--\; △\; -->}BED\sim <!--\; \backslash sim\; -->\mathrm{△<!--\; △\; -->}ABC$. In fact, I think this triangle is congruent to the one mentioned in the last paragraph. Also not sure on that, though.
The last one came about as a surprise while I was working on a proof, but I could not prove it nor could I think of any other ways to construct a similar right triangle to one given. So, my question: given a right triangle, what are some ways to construct a similar triangle (preferably ones that you found clever/interesting)?

How do you prove a triangle with the hypotenuse of length 5 and other sides with lengths 3 and 4 is a right triangle?