# Tag Info

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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