13

Consider the expression $$ (\cos A + \cos B + \cos C)^2+(\sin A + \sin B + \sin C)^2\ . $$ Using the relations $\sin^2 x + \cos^2 y=1$ and $\cos(x-y) = \cos x \cos y + \sin x \sin y$, this expression simplifies to $$ 3 + 2 (\cos(A-B) + \cos(B-C)+\cos(C-A))\ . $$ Plugging in the provided value $-\frac{3}{2}$ for the part in parentheses, we find that the ...


11

We can ignore the $\tanh$ component momentarily and consider instead the function $$f(x) = \frac{\sin x + \sinh x}{x}.$$ This has series expansion $$f(x) \sim 2 + \frac{x^4}{60} + \frac{x^8}{181440} + O(x^{12}).$$ Therefore, on the open interval $x \in (-1,1)$, $f(x)$ has exceptionally small deviation from $2$, thus on the real line, $$g(x) = \frac{\sin \...


8

Denote the mid-point of $OM$ be $N$. Then $N$ is the center of the circle passing through the points $O,P,M,K$. We have: $$\angle PNK = 2\alpha,\quad \angle PMK = \pi-\alpha$$ By Cosine Theorem, $$PK^2 = a^2+b^2-2ab\cos \angle PMK = PN^2 + KN^2 - 2PN\cdot KN \cos 2\alpha$$ Writing $PN = KN = \frac12 OM = r$, we have: $$a^2 + b^2 +2ab\cos \alpha = 2r^2(1-\cos ...


3

Maybe it's better to work with the inequality directly. $f$ is increasing if $f' > 0$. So we require $$1 - \frac{1}{\cos^2 x} > 0,$$ or $$\cos^2 x - 1 > 0, \quad \cos x \ne 0.$$ But because $-1 \le \cos x \le 1$, we have $\cos^2 x \le 1$, hence this inequality is never satisfied, and $f$ is never strictly increasing.


3

Reflect $M$ across $OA$ and $OB$ and you get $A'$ resp. $B'$. Let $t=AB$, then $A'B' = 2t$ and $OA'= OB' = OM = m$. Also $\angle A'OB' = 2\alpha$. By the Law of cosine in triangle $ABM$ we have $$ t = \sqrt{a^2+b^2+2ab\cos \alpha}$$ and finally in isosceles triangle $A'B'O$ we have $$\boxed{m = {t\over \sin \alpha}}$$


3

A delightful application of linear algebra: Let $R(\theta)$ be the linear map $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ that rotates points counterclockwise an angle $\theta$ about the origin. It is well-known that $R(\theta)$ has the matrix representation: $$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$ It is ...


3

Let's set $$f:[-\pi,\pi]\to\Bbb R:f(x)=\cos \sqrt{\frac{\pi^2-x^2}3} + 2\cos \frac{\pi -x}3$$ The two cosine arguments are roughly equal in size, so it actually turns out that the scalar $2$ is very useful. Indeed, for every $x\in(-\pi,\pi]$, we have that $$\bigg|2\cos \frac{\pi -x}3\bigg|>\bigg|\cos \sqrt{\frac{\pi^2-x^2}3}\bigg|\implies \operatorname{...


3

As I have pointed out in a comment, the mistake comes from the power rule of integration. \begin{align} \int\sqrt{\sin x}\cos^3x\,dx&=\int\sqrt{\sin x}\cos^2x\cos x\,dx\\ &=\int\sqrt{\sin x}(1-\sin^2x)\cos x\,dx\\ &=\int\sqrt u(1-u^2)\,du\\ &=\int\sqrt u\,du-\int\sqrt{u^5}\,du\\ &=\frac23u^{3/2}-\frac27u^{7/2}+C\\ &=\frac23(\sin x)^{3/...


3

What you are searching for is the reference angle $\theta$ which is in the first quadrant. Once we have that, then the candidate angles are $\theta, 180-\theta, 180+\theta,$ and $360-\theta$. But since the trig functions for angles in the first quadrant are always positive, we want to consider the absolute value of the ratio. There is also probably a ...


3

$$0=\cos5x \;\cos3x-\sin3x\;\sin x-\cos 2x$$ $$=\cos5x \;\cos3x-\sin3x\;\sin x-\cos (5x-3x)$$ $$=\sin 3x\sin 5x-\sin 3x\sin x$$ $$\to -2\sin ^2 3x\cos 2x=0$$ Can you finish it? Note that $\cos 2x\neq 0$ has to be included(why?) $$\sin 3x=0$$


3

$$\cos (5 x) \cos (3 x)-\sin (3 x) \sin (x)=\cos (2 x)$$ $$\frac{1}{2} (\cos (2 x)+\cos (8 x))-\frac{1}{2} (\cos (2 x)-\cos (4 x))=\cos(2x)$$ $$\frac{1}{2} (\cos (4 x)+\cos (8 x))-\cos(2x)=0$$ $$\cos (8 x)+\cos (4 x)-2 \cos (2 x)=0$$ $$2 \cos (2 x) \cos (6 x)-2 \cos (2 x)=0$$ $$\cos(2x)\left[\cos(6x)-1\right]=0$$ $$\cos(2x)=0\to x=\pm\frac{\pi}{4}+k\pi$$ ...


3

Hint: $$\sqrt{100n^2+n+1}=10n\sqrt{1+\dfrac{n+1}{100n^2}}\sim10n+\dfrac{n+1}{20n}\to10n+\frac1{20}.$$ So the question amounts to finding the cotangent of $\dfrac\pi{20}=9°$. The following table can be very useful: https://math.la.asu.edu/~surgent/mat170/Exact_Trig_Values.pdf.


2

If the new triangle formed is also right (that is, the new $215$ side is parallel to the old $260$ side), then the small triangle is similar to the large one. The corresponding dimensions are in proportion. The smaller triangle is scaled down to $\frac {215}{260}$ of the large one. So if the original base (large) was $161$, the new one (small) is...? Can you ...


2

Let $m=OM$, then $${a\over m} = \sin \alpha _1$$ and $${b\over m} = \sin \alpha _2$$ so $$\cos \alpha _1 = \sqrt{1-{a^2\over m^2}}\;\;\;\;\;\;\;\;\;\; \wedge \;\;\;\;\; \;\;\;\;\; \cos \alpha _2 = \sqrt{1-{b^2\over m^2}}$$ so $$\sin \alpha = {a\over m}\sqrt{1-{b^2\over m^2}} + {b\over m} \sqrt{1-{a^2\over m^2}}$$ Now you have to find out $m$. :)


2

Let $d=OM$ $$ \arcsin(a/d)+ \arcsin(b/d) = \alpha$$ $$ a \sqrt{d^2-b^2} + b \sqrt{d^2-a^2} = d^2\sin \alpha$$ Let Mathematica solve fourth order equation for $d$ $$sal= \sin \alpha$$ {d -> 0.5` Sqrt[(4.` (1.` + 1.` b^2))/sal^2 + 2.` Sqrt[(4.` (1.` + 1.` b^2)^2)/sal^4 - ( 16.` (0.25` - 0.5` b^2 + 0.25` b^4 + 1.` b^2 sal^2))/sal^4]]}


2

A two-liner : Since $OPMK$ is cyclic, $\angle PMK = \pi - \angle POK = \pi - \alpha$; by cosine-rule in $\triangle PMK$, $PK^2=a^2+b^2+2ab\cos \alpha$. Now $OM$ is simply the diameter of the circle which can be found by sine-rule in either of $\triangle POK$ or $\triangle PMK$, $$OM=2r=\frac{PK}{\sin \angle POK}=\frac{PK}{\sin \alpha}=\frac{\sqrt{a^2+b^2+...


2

$A=0$ because in the product there is the factor $\cos 90°=0$.


2

Consider that you look for the zero of function $$f(x)=2^{x} - 2^{-x} - 2\cos \left(\frac{x}{5}\right)$$ By inspection, you know that the solution is somewhere between $x=1.0$ and $x=1.5$ and the function is quite close to a straight line (this is good for root-finding methods). So, be as lazy as I am and start Newton methods with $x_0=0$. It will give the ...


2

$$X^5 = -1 = e^{(2k+1)\pi i}$$ $$X = e^{\frac{(2k+1)}{5}\pi i}, \quad (k=0,1,...,4)$$ $$X \in \{ e^{\pi i /5}, e^{3\pi i /5}, e^{5\pi i /5}, e^{7\pi i /5}, e^{9\pi i /5} \}$$ As you wrote, apply De Moivre's theorem.


2

Form the equation of the line. Then the question on which side of the line a point is can be answered by calculating the value of the defining (linear) function of the line. On the line its value will be $0$. The value is positive/negative if the point is on the left/right (when looking to the direction of the vector defining the line). In this case we get ...


2

yes it can, so lets use your assumption that: $$y=a\sin(bx)$$ however there will be an associated error for each term, namely: $$r_i=y_i-y(x_i)$$ $$r_i=y_i-a\sin(bx_i)$$ now you simply define your sum: $$S=\sum r_i^2=\sum\left[y_i-a\sin(bx_i)\right]^2$$ now you want to optimise for $a,b$ such that $S$ is as small as possible, which would mean: $$\frac{\...


2

Observe that $$\sec^2\theta=\frac1{\cos^2\theta}=1+\tan^2\theta$$ so $$x\tan\theta+\frac{gx^2}{2v^2\cos^2\theta}=x\tan\theta+\frac{gx^2}{2v^2}\left(1+\tan^2\theta\right)\implies$$ our equation is $$\frac{gx^2}{2v^2}\tan^2\theta+x\tan\theta+\frac{gx^2}{2v^2}-y=0$$ The above is a quadratic in $\;\tan\theta\;$ whose discriminant is $$\Delta=x^2-4\frac{gx^2}{2v^...


2

Just for fun: $$ [-\pi/ 4, \pi/ 4] \ni \phi \mapsto c(\phi) = \left( \frac 12 - \frac{\sin \phi}{\sqrt 2},\frac 12 + \frac{\sin \phi}{\sqrt 2} \right) $$ is a parametrization of the diagonal from $(1, 0)$ to $(0, 1)$ in the unit square $[0, 1]^2$, and its length is $$ \sqrt 2 = \int_{-\pi/4}^{\pi/4} \Vert c'(\phi) \Vert \, d\phi = \int_{-\pi/4}^{\pi/4} \...


2

Let $t:=\tan x$ so $\tan y=\frac{b}{at}$ and we want$$\frac{1+t^2}{at^2+b}+\frac{1+b^2/(a^2t^2)}{b^2/(at^2)+b}=\frac{b(1+t^2)+at^2+b^2/a}{b(at^2+b)}=\frac{a+b}{ab},$$taking out a factor of $t^2+b/a$.


2

Almost there. The area under the first half-period is $2$. You're therefore looking for the constants $a_i$ such that $$\int_{a_i}^{a_{i+1}} \sin x dx = \frac2n$$ The integral is indeed (negative) cosine: the left-hand side is $$-\cos x\big|_{a_i}^{a_{i+1}} = -\cos(a_{i+1}) + \cos(a_i)$$ Now, the first segment starts at $a_0 = 0$, so we have $$-\cos(a_1) + \...


2

Hint & caution: If the load is constant (and just resistive), the instantaneous power is proportional to the square of the sine wave, and the power average over a period will be proportional to the integral of the $\sin^2$. Additional note: If you intend to use a modern led lamp, then you have to radically change your approach!


2

The problem has no solutions because there is no $\theta$ such that $\tan\theta=\frac{\sqrt{15}}3$ and that $\cos\theta=\frac{\sqrt{10}}4$. In fact, if $\cos\theta=\frac{\sqrt{10}}4$, then$$\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=\frac{1-\cos^2\theta}{\cos^2\theta}=\frac{\frac6{16}}{\frac{10}{16}}=\frac35,$$and therefore $\tan\theta=\sqrt{\frac35}$.


2

Why isn't this true for the function $f(x)=x^2$? We have $f(0)=0^2=0$ and $f(1)=1^2=1$, so shouldn't $f(1/2)=(1/2)^2$ be in the middle of these two values? But no, $f(1/2)=1/4$. The function $f(x)=x^2$ isn't linear: we don't have $f(ax)=a\cdot f(x)$ in general; that is, if you scale $f(x)$ by some number, that's not the same as if you scaled the input to ...


2

Starting from $R=\sqrt{65}$, $a=\arcsin 7/\sqrt{65}$, we have $$\sqrt{65}\sin(x+a)=6$$ $$\Rightarrow x=\arcsin \frac{6}{\sqrt{65}}-a=\arcsin \frac{6}{\sqrt{65}}-\arcsin \frac{7}{\sqrt{65}}$$ Using $$\arcsin u - \arcsin v=\arcsin (u\sqrt{1-v^2}-v\sqrt{1-u^2})$$ $$x=\arcsin\left(\frac{6}{\sqrt{65}}\cdot \frac{4}{\sqrt{65}}-\frac{7}{\sqrt{65}}\cdot \frac{\sqrt{...


2

You have: $2\tan(\frac{A}{2}) = \dfrac{\sin A}{\sin B\sin C}\implies 2\dfrac{\sin(\frac{A}{2})}{\cos(\frac{A}{2})}= \dfrac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin B\sin C}\implies \sin B\sin C = \cos^2(\frac{A}{2})= \dfrac{1+\cos A}{2}\implies 1+\cos A = 2\sin B\sin C= \cos(B-C)-\cos(B+C)= \cos(B-C)+\cos A\implies \cos(B-C) = 1\implies B-C=0\implies B = C$....


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