# Tag Info

14

This is what I'm trying to show on a diagram indicating explicitly all quantities and the approximation is quite rough. Here is a circle with center $O$. Quantities $OA=OB=OC=OD=AB=1$ and $OC\perp OA, OD\parallel OA$. Line segments $AC$ and $BD$ have an intersection $E$. We can easily deduce the following quantity. $$AC^2=OA^2+OC^2=2\qquad ... 8 Riffing on @Shuchang's answer ... Starting with unit circle \bigcirc O, one easily constructs A, B, C, D with |\overline{AB}| = \sqrt{2} and |\overline{CD}| = \sqrt{3}. Quadrisecting \overline{AB} and \overline{CD} one draws \bigcirc A and \bigcirc{B} to provide chords of length \sqrt{2}/4 and \sqrt{3}/4. Chains of congruent ... 6 Here is the integral with the Weierstrass substitution:$$ \begin{align} \int\frac{\mathrm{d}x}{1+\sin(x)} &=\int\frac1{1+\frac{2z}{1+z^2}}\frac{2\,\mathrm{d}z}{1+z^2}\\ &=\int\frac{2\,\mathrm{d}z}{1+2z+z^2}\\ &=\frac{-2}{1+z}+C\\ &=\frac{-2}{1+\tan(x/2)}+C \end{align} $$so your answer is correct. Now consider$$ \begin{align} ...

5

Note that $\sqrt 2 \sin^2 x = \sqrt 2(1 - \cos^2 x)$. Make that substitution, \begin{align} \sqrt 2 \sin^2 x + \cos x = 0 & \iff \sqrt 2(1-\cos^2 x) + \cos x = 0\\ \\ &\iff 1 - \cos^2 x + \frac 1{\sqrt 2}\cos x = 0\\ \\ &\iff \cos^2 x - \frac 1{\sqrt 2}\cos x - 1 = 0\end{align} and put $y = \cos x$...You'll have a quadratic equation in ...

5

There is no need for complicated Cauchy products. We will simply use $\Big(f^2(x)\Big)'=2f(x)f'(x)$. $$\sin(x)=\sum_0^\infty(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\iff\sin'(x)=\sum_0^\infty(-1)^{n}\frac{x^{2n}}{(2n)!}=\cos(x)$$ $$\cos(x)=1+\sum_1^\infty(-1)^{n}\frac{x^{2n}}{(2n)!}\iff\cos'(x)=\sum_1^\infty(-1)^{n}\frac{x^{2n-1}}{(2n-1)!}=-\sin(x)$$ Now, ...

5

You don't need the double angle formula to solve this one! $\cos(3\theta)=1/2$ $3\theta=\cos^{-1}(1/2)$ $\cos^{-1}(1/2)=60$ (in degrees) $3\theta=60$ $\theta=20$ but hang on a second!! there are more solutions within $0\leq \theta\leq 360^\circ$. ... actually this domain must be modified for $3\theta$ since you need the answer to $3\theta$ not just ...

4

The simplest way of deriving/proving this is through Euler's formula: Take $e^{i(\theta+\alpha)}$ which, through basic exponent rules we know is equal to $e^{i\theta +i\alpha} = e^{i\theta}\cdot e^{i\alpha}\ .$ Now expanding this (using Euler's formula) we get: \begin{align} e^{i\theta}\cdot e^{i\alpha} &= ... 4 Recall the inverse tangent identity\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x+y}{1-xy}.$$So in particular,$$2 \tan^{-1}x = \tan^{-1} \frac{2x}{1-x^2},$$and$$3 \tan^{-1} x = \tan^{-1} \frac{x(3-x^2)}{1-3x^2}.$$Consequently, we wish to find x such that$$\frac{x(3-x^2)}{1-3x^2} = -\frac{10}{9 \sqrt{3}},$$which is equivalent to$$0 = 9 \sqrt{3} ...

4

Let's start with our original expression: $$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}$$ We need to prove that: $$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\cos^2 x$$ Step 1: Recall that $\tan x = \dfrac{\sin x}{\cos x}$. This means that $\tan^2 x=\dfrac{\sin^2 x}{\cos^2 x}$. $$\dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x}=\dfrac{\cos^2 x - \sin^2 ... 4 Convert into a polynomial of \cos(40^\circ):$$ \begin{align} &\cos^2(10^\circ)+\cos^2(50^\circ)-\sin(40^\circ)\sin(80^\circ)\\ &=\sin^2(80^\circ)+\sin^2(40^\circ)-\sin(40^\circ)\sin(80^\circ)\tag{1}\\ &=4\sin^2(40^\circ)\cos^2(40^\circ)+\sin^2(40^\circ)-2\sin^2(40^\circ)\cos(40^\circ)\tag{2}\\ ...

3

Taking a look at it specifically. If $$cos(\theta) = \frac{-4}{5}$$ That means the hypotenuse of your triangle is $5$ and the adjacent length is $4$ (on the negative x-axis) Using Pythgorean's theorem: you get $a^2 = 25 - 16$ $$a^2 = 9$$ $$a = \pm 3$$ Since the angle $\theta$ lies between $90^{{\circ}}$ and $180^{{\circ}}$ that means your triangle lies in ...

3

No, $\sin (x) \sin (y) \neq \sin (xy)$ in general. $\sin 2 \approx 0.9093$ and $\sin 4 \approx -0.7568 \neq 0.9093^2$ There is an identity $\sin(x)\sin(y)=\frac 12(\cos(x-y)-\cos(x+y))$ but you may not find that an improvement. If you set $x=y$ in this you get $\sin^2x=\frac 12(1-\cos(2x))$

3

For the partial products we have \begin{align} \prod_{k = 0}^N \cos \frac{x}{2^k} &= \frac{\sin \frac{x}{2^N}\cos\frac{x}{2^N}}{\sin \frac{x}{2^N}} \prod_{k=0}^{N-1} \cos \frac{x}{2^k}\\ &= \frac{\sin \frac{x}{2^{N-1}}}{2\sin \frac{x}{2^N}} \prod_{k=0}^{N-1} \cos \frac{x}{2^k}\\ &= \frac{\sin \frac{x}{2^{N-1}}\cos \frac{x}{2^{N-1}}}{2\sin ... 3 Well, if you buy or have access to (through previous learning) e^{i\theta} = \cos \theta + i \sin \theta, then setting \theta = a - b one has e^{i(a - b)} = \cos(a - b) + i\sin(a - b). \tag{1} But also, e^{i(a - b)} = e^{ia}e^{-ib} = (\cos a + i \sin a)(\cos b - i\sin b). \tag{2} The imaginary part of the product on the right of (20) is \sin a ... 2e^{i\alpha}=\cos \alpha + i\sin \alphae^{i\theta}=\cos \theta + i\sin \theta\begin{align}e^{i(\alpha+\theta)}&=e^{i\alpha}\cdot e^{i\theta} \text{ (law of exponents)}\\ \cos(\alpha+ \theta) + i\sin (\alpha+\theta)&=(\cos\alpha+i\sin\alpha)(\cos \theta + i\sin \theta) \end{align}$$Multiply, and compare \Re(z) and \Im(z) of both ... 2 With vectors there is a geometric formula for the dot product$$\vec{u}\cdot\vec{v}=\left\|\vec{u}\right\|\left\|\vec{v}\right\|\cos\theta$$where \theta is the angle between the two vectors. Take \vec{u}=\langle\cos(a),\sin(a)\rangle and \vec{v}=\langle-\sin(b),\cos(b)\rangle and this formula gives$$\sin(a)\cos(b)-\cos(a)\sin(b)=\cos\theta$$... 2 From$$e^{ix} = \cos{x} + i\sin{x}$$we can derive that$$\sin{x} = \frac{e^{ix} - e^{-ix}}{2i}$$and$$\cos{x} = \frac{e^{ix} + e^{-ix}}{2}$$Hence, we have$$\sin{a}\cos{b} - \sin{b}\cos{a} = \frac{(e^{ia} - e^{-ia})(e^{ib} + e^{-ib})}{4i} - \frac{(e^{ib} - e^{-ib})(e^{ia} + e^{-ia})}{4i}\\ = \frac{e^{i(a+b)} - e^{i(-a+b)} + e^{i(a-b)} - e^{i(-a-b)} - ...

2

You could use this $$e^{\sin(x)} = e^{\frac{1}{2i}(e^{ix} - e^{-ix})}$$ But I doubt that's what you're looking for. A common trick in situations like this is to use the taylor series. So you could try $$e^{\sin(x)} = e^{x-x^3/6 + x^5/120 - \cdots}$$ or $$e^{\sin(x)} = 1 + \sin(x) + \sin^2(x)/2 + \sin^{3}(x)/6 + \cdots$$

2

$$\text{Let} \cos^{-1}0.8 = \theta\\ \cos \theta = 0.8 \\ 1-2\sin^2\frac{\theta}{2} = 0.8 \\ \sin^2\frac{\theta}{2} = 0.1 \\ \sin\frac{\theta}{2} = \sqrt{0.1} \space \text{(think about why it is not negative)} \\ \sin\left(\frac{1}{2}\cos^{-1}0.8\right) = \sqrt{0.1}$$ Hope you can do the other one too. (Use the same approach)

2

Using the identity $$\cos(x2^{-k})=\frac{\sin(x2^{1-k})}{2\sin(x2^{-k})}$$ we get \begin{align} \prod_{k=0}^n\cos(x2^{-k}) &=\prod_{k=0}^n\frac{\sin(x2^{1-k})}{2\sin(x2^{-k})}\\ &=\frac1{2^{n+1}}\frac{\prod\limits_{k=-1}^{n-1}\sin(x2^{-k})}{\prod\limits_{k=0}^n\sin(x2^{-k})}\\ &=\frac1{2^{n+1}}\frac{\sin(2x)}{\sin(x2^{-n})}\\[12pt] ... 2 I can only see a relation with trigonometric functions, i.e.something like \sqrt{3}=2-\tan (\pi/12) and \sqrt{2}=\frac{4\cos(\pi/12)}{3-\tan(\pi/12)}. $$If we consider the continued fraction of \sqrt{2}, \sqrt{3} and 4/\pi we see some similarities, too. However, there are also arguments indicating that both numbers are only accidentally close. ... 2 Series are absolutely convergent, so we can expand using Cauchy products:$$\begin{align}\cos^2 x&=\left(\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}\right)^2=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{n-k}\frac{x^{2n-2k}}{(2n-2k)!}(-1)^k\frac{x^{2k}}{(2k)!}\\&=\sum_{n=0}^\infty x^{2n}\sum_{k=0}^n\frac{(-1)^n}{(2n-2k)!(2k)!}=1+\sum_{n=1}^\infty ...

2

So, to see that $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$, see http://www.math.wisc.edu/~leili/teaching/math222s11/problems/quizzes/trig.pdf. Then, use the fact that $\cos(-b)=\cos(b),\sin(-b)=-\sin(b)$. So, $$\cos(a-b)=\cos(a+(-b))=\cos(a)\cos(-b)-\sin(a)\sin(-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$$ Q.E.D.

2

In case triangle it is useful to draw our right triangle: Now let us remember the trig definitions for any given angle $\theta$: \begin{align} \sin(\theta) &= \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}\ \ &\csc(\theta) &= \frac{\mathrm{hypotenuse}}{\mathrm{opposite}}\\ \cos(\theta) &= \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} ... 2 Well, it's rather simpled = \sqrt{a^2 + b^2 - 2ab \ cos(\theta)}d^2 = a^2 + b^2 - 2ab \ cos(\theta) d^2 - a^2 - b^2=2ab \ cos(\theta) \frac{d^2 - a^2 - b^2}{2ab} = \ cos(\theta) \frac{d^2}{2ab} - \frac{a^2 + b^2}{2ab} = \ cos(\theta)$$and since a and b are fixed, k = \frac{a^2 + b^2}{2ab}$$ \ cos(\theta) = \frac{d^2}{2ab} -k$$... 2 Yup: it states that$$ a=\sqrt{b^2+c^2-2bc\cos\alpha} $$in stantard trigonometric notation (where a,b,c are the sides of the triangle, and \alpha is the angle which is opposite to a). A simple proof can be given with vector calculus: being \vec{a}=\vec{b}-\vec{c}, squaring this relation you obtain:$$ \vec{a}\cdot\vec{a}\equiv a^2 = b^2 + c^2 - 2 ...

1

Sounds like an approach to your problem is to let $a=b$ for simplicity so $$d = \sqrt{a^2+b^2-2ab \cos x} = a \sqrt{2-2\cos x}$$ and then just plot the function $f(x) = \sqrt{2-2\cos x}$, getting which is clearly a wave, but not sure how sinusoidal it is. It is, however, similar to a similar transformation of the sine, just off by a horizontal ...

1

The function that describes the distance of a fixed point from a point travelling on a circle cannot be perfectly "sinusoidal", since it exhibits discontinuities for the derivatives, as shown by @gt6989b, for instance. However we can state that the coefficients of the Fourier cosine series of $$f(\theta) = \sqrt{1-\lambda \cos\theta},$$ where we set ...

1

Square it. You get $$1 - 2 \sin x \cos x = \sin^2x + \cos^2 x - 2 \sin x \cos x = (\sin x - \cos x)^2 = 1$$ which implies $\sin x = 0$ or $\cos x = 0$. Therefore the only possible solutions are multiples of $\pi/2$. See which ones are actual solutions of your problem (we squared, so $\sin x - \cos x = \pm 1$ when $x = k \pi / 2$). (Only $3 \pi / 2$ works ...

1

Rewrite the equation as $\sin x=\cos x-1$, square both sides, and use the identity $\sin^2x=1-\cos^2x$ to get $$1-\cos^2x=\cos^2x-2\cos x+1$$ which simplifies to $$\cos x(1-\cos x)=0$$ so that either $\cos x=0$ or $\cos x=1$. The former corresponds to $x=\pi/2$ and $3\pi/2$ in the interval $0\lt x\lt2\pi$. The latter corresponds to $x=0$ or $2\pi$, ...

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