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Start by drawing a picture of where is the light with respect to the walls. It will help you understand the problem. At $t=0$, the light points toward one of the walls. As it starts to rotate, the lig …

The entire triangle is below the $x$-axis, so your center is $(1/3,-1/3)$. If you move the center, you need to add $(-1/3,1/3)$ to every coordinate. I assume that you don't rotate the triangle, so the …

The following identities were used: $$\tan(\theta/2)=\frac{\sin(\theta/2)}{\cos(\theta/2)}\\\sin^2(\theta/2)+\cos^2(\theta/2)=1\\|e^{i\phi}|=1$$ The solution is not unique. Multiplying both $a$ and $b …

You can write this equation as $$f(t)=A\sin(2\pi f t)$$ The relationship between frequency of the light $(f)$, wavelength $(\lambda)$, and speed $(c)$ is $$c=\lambda f$$ From here $$f=\frac{c}{\lambda …

You use the second form for the displacement as a function of time $$x=A'\sin(bt+B')$$ The velocity is then $$v=A'b\cos(bt+B')$$ and the acceleration is $$a=-A'b^2\sin(bt+B')$$ The ratio of accelerat …

It would be much easier for you if you would make a sketch. Let's use three points on a 2D grid (marked with North, South, East, and West). The first point, say $(x_1,y_1)$ is your ship. The enemy is …

You get the same position in the unit circle if you add multiples of $360^\circ$. Second quadrant means angles between $90^\circ$ and $180^\circ$. So take a look at several $a$ values: $a=0$ means $\ …

With the correction pointed out by @Lozenges, you have a sum of square numbers that is equal to $0$. That is true only if all terms are $0$. The solution is then given by $$\sin x=\cos y=\pm 1$$ You c …

Use common denominator, then expand anything with $+$ sign in a trigonometric function: $$\cos\left(\frac12nx\right)\;\sin\left(\frac12(n+1)x\right)-\sin\left(\frac12x\right)=\cos\left(\frac12(n+1)x\r …

Let $x$ and $b$ be some vectors of length $N$, and $A$ an $N\times N$ matrix. Let's write the original curve in terms of elements: $$x^TAx+b^Tx+c=0\\\sum_{i=1}^N\sum_{j=1}^Na_{i,j}x_i x_j+\sum_{i=1}^N …

Both have to climb the 28 degree slope. So what they do is instead of going straight up, they go sideways. This is a 3D geometry problem, not a 2D one. I think this should be enough of a hint Dotte …

You can use simple vector addition. Say $$\hat a=\frac{\vec{OA}}{|\vec{OA}|}\\\hat b=\frac{\vec{OB}}{|\vec{OB}|}$$ Since $|\hat a|=|\hat b|=1$, $\hat c=\hat a+\hat b$ points along the bisector of $\an …

The angle between the red line and the blue line in $30^\circ$. Then the ratio of the blue length to red length is $\cos 30^\circ$. So $$red=\frac{blue}{\cos 30^\circ}\approx 1.15\ blue$$

If $$x=\arctan\frac 12+\arctan\frac 15+\arctan\frac 18$$ you can take the tangent of both sides: $$\tan x=\tan\left(\arctan\frac 12+\arctan\frac 15+\arctan\frac 18\right)$$ Then you will need to find …

You can write the equation as $y\cot y=1$. One of the solutions is $y=0$. The rest can only be found numerically. For $y=0$ you get $q=1$