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The double angle formulas follow by letting a = b in each of the trig. expansion identities for f(a+b).

A derivation based on the unit circle may be found in the Year 11 3U Pender (Cambridge) textbook. (Also you should be able to see that some of them follow immediately from the others once you've derived those, using properties of the trig. functions.)

You needed to take the negative square root, since y needed to be ≤ 2.

$\noindent Let its displacement at time $t$ (measured in seconds) be $x\equiv x(t)$ (measured in metres), so we are given $\ddot{x}=12$ (constant). Then integrating yields $\dot{x}=c_1 + 12t$ for some constant $c_1$. When $t=5$, $\dot{x}=2$, so $2=c_1 + 60 \Rightarrow c_1 = -58$. Thus $\dot{x} =...

Re: First Year Uni Calculus Marathon Going to infinity counts as "exist" for the purposes of using L'Hôpital's rule. You can't have something that oscillates forever though (like (2+cos(x))/(2-cos(x)) or something), that'd fail to satisfy the requirement for L'Hôpital's rule.

$\noindent The parabola is defined as the set of all points for which the distance to the focus equals the distance to the directrix (or equivalently the squared-distances are equal). Let $P=(x,y)$, then its squared-distance to the focus is $(x-2)^2 + (y-4)^2$. Also, its distance to the line...

$\noindent Note that $67.5^\circ$ is half of $135^\circ$, and we know the exact value of $\cos \left(135^\circ\right)$. So we can apply the half-angle identities to find the exact values, noting to take the positive square root since the angles are acute.$ $\noindent The half-angle identities...

You have found a line that is perpendicular to one of the given lines, but this isn't what we want. The lines we want are essentially the lines containing all those points that are equidistant from the two given lines (they're equidistant because the lengths of the perpendiculars from P to the...

Draw the two lines going through the point of intersection of the given lines and bisecting the angles made by these two given lines. This is the locus.

Another way to do it is to recall that 2t/(1 – t2) = tan(theta), where t = tan(theta/2). In this case, theta/2 = 120 deg, so theta = 240 deg, so the expression equals tan(240 deg) = tan(60 deg) = sqrt(3).

The locus will be two lines. Both cases need to be considered to get the full locus.

It's the pair of intersecting lines that bisect the angles made by the two given lines (they go through the point of intersection of those two lines). If you want the equations of the lines, the way they'd expect you to do it is to use the "perpendicular distance formula" and then equate the...

It's just sqrt(3).

Try just drawing a perpendicular line segment from a random point P to one of the lines, and then to the other line (image: https://www.mathsisfun.com/images/perpendicular.gif ).

$\noindent Draw from $P$ line segments to the given lines that are perpendicular to those lines. The points where they meet the given lines are $M$ and $N$. (The answer to the locus will be the lines bisecting the angles made by the lines.)$

Re: MATH1131 help thread $\noindent So for example, let's set up how to find the intersection with the $xz$-plane. This is where $y=0$, so we're looking for a point on the line with $y=0$, i.e. we want $ \begin{bmatrix} 1 \\ 2 \\ -3\end{bmatrix} + t\begin{bmatrix}2 \\ 3 \\ 5\end{bmatrix}=...

Re: MATH1131 help thread This question is asking for when it intersects the coordinate planes, rather than coordinate axes, which is what you originally asked for. But it's a similar idea. To find the place where it cuts the xy-plane for instance, we'd set z = 0 (because the xy-plane is the...

Re: First Year Uni Calculus Marathon $\noindent Let $P$ and $Q$ be non-constant polynomials with positive leading coefficients. Show that $\lim _{y\to \infty}\frac{P\left(y\right)}{Q\left(\ln y\right)}=\infty$.$

Re: MATH1131 help thread To find the intercept with one of the axes, set the other two variables to 0 and try and solve for the variable whose axis' intercept we're finding. (If you get no solution, it means the line doesn't intersect that axis. A line in 3D need not have any axial intercepts...

Re: MATH1131 help thread $\noindent If the line is $\frac{x-a}{v_1}=\frac{y-b}{v_2}=\frac{z-c}{v_3}$, then the line is $\begin{bmatrix}x\\y\\z\end{bmatrix}= \begin{bmatrix}a\\b\\c\end{bmatrix}+ t \begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}$.$ $\noindent To see this, since each of those...