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10

WLOG $\displaystyle\alpha=90^\circ\implies\beta+\gamma=90^\circ\iff\gamma=90^\circ-\beta$ $$\sin^2\beta+\sin^2\gamma=\sin^2\beta+\sin^2(90^\circ-\beta)=\sin^2\beta+\cos^2\beta=?$$ $$\cos^2\beta+\cos^2\gamma=?$$


10

Taking exponential of both sides, you want $$ \prod_{k=1}^n \left(2 \cos\left(\dfrac{2\pi\cdot 3^k}{3^n+1}\right)+1\right) = 1 $$ Now note that $1 + 2 \cos(x) = \sin(3x/2)/\sin(x/2)$ and the product telescopes to become $$\dfrac{\sin \left( \dfrac{3^{n+1}\pi}{3^n+1}\right)}{\sin \left( \dfrac{3 \pi}{3^n+1}\right)} = \dfrac{\sin \left( 3\pi - ...


7

Recall that $\sin(a-b)=\sin a\cos b -\cos a\sin b$. Let $a=x+h$ and $b=x$. Remark: You used precisely the same approach, except that instead of using $\sin(x+h-x)$, as above, you used $\sin(x+h-h)$.


7

Rewrite as $$\int(1-\cos^2 x) \cos^2 x \sin x \ dx$$ and let $t=\cos x\Rightarrow dt = -\sin x \ dx$.


5

In a right triangle, exactly one of $\alpha$, $\beta$, $\gamma$ must equal $\pi/2$, so let this be $\gamma = \pi/2$. Then $\sin \gamma = 1$ and $\cos \gamma = 0$. We must also have $\alpha + \beta = \pi/2$, hence $\beta = \pi/2 - \alpha$ and it immediately follows that $\sin \beta = \sin (\pi/2 - \alpha) = \cos \alpha$, and $\cos \beta = \sin \alpha$. The ...


5

The fastest way is probably to use the identity JinnyK4542 suggested: $\sin(x) + \sin(360^\circ - x) = 0$. Here's an alternative which can be more easily generalized: We want to find $$S = \sin\frac{\pi}{18} + \sin\frac{2\pi}{18} + ... + \sin\frac{36\pi}{18}$$ Consider $e^{i\theta} = \cos\theta + i \sin\theta$. We have $$\begin{align}S &= ...


4

Consider $m$ number of $\dfrac{\sin^2\alpha}m$ and $n$ number of $\dfrac{\cos^2\alpha}n$ As each of the term $\ge0$ for real $\alpha;$ using AM, GM inequality $$\frac{m\cdot\dfrac{\sin^2\alpha}m+n\cdot\dfrac{\cos^2\alpha}n}{m+n}\ge \left[\left(\dfrac{\sin^2\alpha}m\right)^m\left(\dfrac{\cos^2\alpha}n\right)^n\right]^{\dfrac1{m+n}}$$


4

The period of $\displaystyle\sin(ax+c)=\frac{2\pi}a$ that of $\displaystyle\cos(bx+d)=\frac{2\pi}b$ Now if $\displaystyle\frac{\dfrac{2\pi}a}{\dfrac{2\pi}b}=\frac ba$ is rational, the period of $\displaystyle\sin(ax+c)+\cos(bx+d)$ will be lcm $\displaystyle\left(\frac{2\pi}b,\frac{2\pi}a\right)$ or its divisor


4

Using the formula $$\cos{(a+b)}=\cos{(a)} \cos {(b)}-\sin{(a)} \sin{(b)}$$ we get the following: $$\cos{ \left ( \frac{3 \pi}{2}+x \right )}=\cos{ \left ( \frac{3 \pi}{2}\right )} \cos{(x)}-\sin{ \left ( \frac{3 \pi}{2} \right )} \sin{(x)}=0 \cdot \cos{(x)}-(-1) \cdot \sin{(x)}=\sin{(x)}$$ $$\cos{(2 \pi+x)}=\cos{(2 \pi)} \cos{(x)}-\sin{(2 \pi )} ...


4

L.H.S. $$\large\frac{\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}}{\cos^2{\alpha}+\cos^2{\beta}+\cos^2{\gamma}}$$ Since, $\alpha+\beta+\gamma$=$180^\circ$ Since, the triangle is a right angled triangle (assuming it is at $\gamma$), that means $\alpha+\beta=90^\circ$ Or, $\beta=90^\circ-\alpha$ So, $\sin \beta$=$\cos \alpha$ and $\cos ...


4

Using complex exponential relations, $$e^{i\theta} = \cos\theta + i \sin\theta \qquad \sin\theta = \frac{1}{2i}(e^{i\theta}-e^{-i\theta}) \qquad \cos\theta = \frac{1}{2}(e^{i\theta}+e^{-i\theta})$$ we see that sine-cosine polynomial of degree $p$ can be written as a linear combination of complex exponentials "as large as" $e^{ip\theta}$ and "as small as" ...


4

Oh my. We have: $$1+i\sqrt{3} = 2\exp\left(i\arctan\sqrt{3}\right)=2\exp\frac{\pi i}{3}$$ $$\frac{1}{1-i} = \frac{1}{2}(1+i) = \frac{1}{\sqrt{2}}\exp\frac{\pi i}{4},$$ hence: $$z=\frac{1+i\sqrt 3}{1-i} = \sqrt{2}\exp\frac{7\pi i}{12},$$ so: $$ z^{40} = 2^{20}\exp\frac{70\pi i}{3}=2^{20}\exp\frac{4\pi i}{3}=-2^{20}\exp\frac{\pi i}{3}=-2^{19}(1+i\sqrt{3}).$$ ...


3

Let $\zeta = \exp(2\pi \textrm{i}/n)$ be a primitive root of unity. Then your sum equals the real part of $$s = \sum_{k=1}^{\frac{n-1}{2}} \zeta^k.$$ Now $$s + \overline{s} + 1 = \sum_{k=0}^{n-1} \zeta^k = 0$$ and $\textrm{Re}(s) = \textrm{Re}(\overline{s})$ so $2 \textrm{Re}(s) + 1 = 0$.


3

Here is another approach. Write the integral as $$I = {2}\int_{0}^{\infty} \frac{\sin^2(x)}{x^2}. $$ Recalling the Mellin transform of a function $f$ $$ \int_{0}^{\infty} x^{s-1} f(x)dx $$ our integral is the Mellin transform of $\sin(x)^2$ with $s=-1$. The Mellin transform is $\sin(x)^2$ given by $$ -\frac{1}{2}\,{\frac {\sqrt {\pi }\,\Gamma ...


3

To show this is true for any polynomial of degree $n+m$, it's sufficient to show that this is the case for an arbitrary term $c^ns^m$ - since every term must be of this form for some $n,m$ - and to check that adding such terms together won't affect the derivative. Note that $$\frac{d}{dx}(c^n s^m)=mc^{n+1}s^{m-1}-nc^{n-1}s^{m+1}$$ This gives another degree ...


3

Rewrite as $a^3+b^3+c^3=(a+b+c)c^2$ and then as $a^3+b^3=(a+b)c^2$ and then as $c^2=a^2-ab+b^2$. Now recall the Cosine Law $$c^2=a^2+b^2-2ab\cos(C).$$


3

Let $\theta$ be an angle which satisfies $\cos \theta = \dfrac{2}{\sqrt{13}}$ and $\sin \theta = -\dfrac{3}{\sqrt{13}}$. Then we have: $2\sin x - 3\cos x$ $= \sqrt{13}\left(\dfrac{2}{\sqrt{13}}\sin x - \dfrac{3}{\sqrt{13}}\cos x\right)$ $= \sqrt{13}\left(cos\theta\sin x + \sin\theta\cos x\right)$ $= \sqrt{13}\sin(x+\theta)$. Do you know what the ...


3

Hint: $$\tan\frac\pi4=1\hspace{5mm} ;-){}{}{}{}$$


3

You have the right trig identity. Let $u=\cos(x)$. Then using $\cos(2x)=2\cos^2(x)-1$ we have a quadratic equation $2+2u^2-1=3u$. Now solve this quadratic equation and substitute back $\cos(x)$


3

Hint Plotting the function is a good idea but your plot is not properly scaled. If you are not able to change the length of the $y$ axis, plot $$20\Big(\sin\frac{3x}{4}+\cos\frac{2x}{5}\Big)$$ and you will visually percieve what lab bhattacharjee means in his good answer.


3

$$\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} = 2\sin\frac{x}{2}\cos\frac{x}{2}\frac{\cos\frac{x}{2}}{\cos\frac{x}{2}} = 2\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\cos^2\frac{x}{2} = \frac{2\tan\frac{x}{2}}{\frac{1}{\cos^2\frac{x}{2}}} = \frac{2\tan\frac{x}{2}}{\frac{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}} = ...


3

$$ \tan ^2 {\frac {\theta}{2}} = \frac{\sin^2 {\frac {\theta}{2}} }{\cos^2 {\frac {\theta}{2}}} = \frac{ \left({\dfrac{1 - \cos \theta}{2}}\right)}{\left({\dfrac{1 + \cos \theta}{2}}\right)} \;\; \text{(since $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta= 1 - 2 \sin^2 \theta = 2 \cos^2\theta - 1$ )}$$ Now substituting for $\cos \theta$ and factorising we ...


3

You may try componendo and dividendo : $\dfrac{1-\tan^2\left(\frac\theta2\right)}{1+\tan^2\left(\frac\theta2\right)}=\dfrac{\cos(\alpha)+cos(\beta)}{1+\cos(\alpha) \cos(\beta)} \\ \iff \dfrac{-2\tan^2\left(\frac\theta2\right)}{2} \stackrel{\color{red}{*}}{=} \dfrac{\cos(\alpha)+cos(\beta) - 1-\cos(\alpha)\cos(\beta)}{\cos(\alpha)+cos(\beta) + ...


3

$$\begin{align*} \tan \theta \sin \theta + \cos \theta & \stackrel{\text{def.}}= \frac{\sin^2 \theta}{\cos \theta} + \cos \theta \\ & \stackrel{\text{Pythagoras}}= \frac{1-\cos^2 \theta}{\cos \theta} + \cos \theta \\ & = \frac1{\cos\theta} - \cos \theta + \cos \theta \\ & \stackrel{\text{def.}}= \sec\theta \end{align*}$$ Where we use the ...


3

Let $z=e^{it}$, then $dz=ie^{it}\ dt$ or $dt=\dfrac{dz}{iz}$, and $\cos t=\dfrac{e^{it}+e^{-it}}{2}=\dfrac{z+z^{-1}}{2}$. \begin{align} \int_0^{2\pi}\frac{1}{4+\cos t}dt&=\oint_C\frac{1}{4+\frac{z+z^{-1}}{2}}\cdot\frac{dz}{iz}\\ &=\frac2i\oint_C\frac{1}{z^2+8z+1}dz, \end{align} where $C$ is the circle of unit radius with its center at the origin. The ...


2

Note that your integral can be rewritten as follows: $$\color{blue}{ I = \int^{2\pi}_0 \frac{2}{8+ e^{it}+ e^{-it}} \, \mathrm{d}t = \int^{2\pi}_0 \frac{2 e^{it}}{e^{i 2 t} + 8 e^{it} + 1} \, \mathrm{d}t}$$ Define now $z \equiv e^{it}$, $\mathrm{d}z = ie^{it} \, \mathrm{d}t = i z \, \mathrm{d}t $, so: $$ \color{blue}{I = \frac{2}{i} \int_\gamma ...


2

You can easily prove that $$d:(x,y)\mapsto \arccos\left(\langle x,y\rangle\right)$$ is a metric. Then, $$\forall x,y,z; d(x,y)\leq d(x,z)+d(z,y)$$ Therefore, if $\theta_{xz}+\theta_{zy}\in [0,\frac{\pi}{2}]\subset [-\frac{\pi}{2},\frac{\pi}{2}]$ ...


2

The small-angle approximations for trigonometric functions are based on their Taylor series. Such as: $$\sin x = x - \frac{x^3}{6} +\frac{x^5}{120} - \dots$$ $$\cos x = 1 - \frac{x^2}{2} +\frac{x^4}{24} - \dots$$ In your example, only the first, largest term of the expansion was used. Sometimes people use more, not only for greater accuracy but also ...


2

Using the Werner formulas, $$\begin{align} \frac{-\cos(x-y)-\cos(x+y)}{-\cos(x-y)+\cos(x+y)} &=\frac{-2\cos{x}\cos{y}}{-2\sin{x}\sin{y}}\\[4pt] &=\cot{x}\cot{y} \end{align}$$


2

Even and odd functions Even functions: $f(x)=f(-x)$. Geometrically, this is symmetry about the $y$-axis. Odd functions: $-f(x)=f(-x)$. Geometrically, this is origin symmetry. From these definitions and the graphs of $\sin x$ and $\cos x$, we can see that $\sin x$ is odd, $\cos x$ even. Note: ${\color{red} \sin \color{red}x}$ is red, $\color{blue}\cos ...



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