# Tag Info

36

Using the sum-to-product formula for cosine $$\cos s + \cos t = 2\cos \tfrac{s+t}{2}\cos \tfrac{s-t}{2} = \cos (s+t)$$ this is a great time to do that 45 degree rotation $u = \tfrac{s+t}{2}:, v=\tfrac{s-t}{2}, s+t = 2u$ $$2\cos u\cos v = \cos 2u \hspace{0.25in}\text{or}\hspace{0.25in} \bbox[5px,border:2px solid #F5A029]{2\cos v = \frac{\cos 2u}{\cos ... 32 Radians, unlike degrees, are not arbitrary in an important sense. The circumference of a unit circle is 2\pi; an arc of the unit circle subtended by an angle of \theta radians has arc length of \theta. With these 'natural' units, the trigonometric functions behave in a certain way. Particularly important is$$\lim_{x\to 0} \frac{\sin x}{x} = 1 ...

21

It works out precisely because $$\lim_{x \to 0} \frac{\sin x}{x} = 1,$$ which in turns happens precisely because we've chosen our angle to be the same as the arclength around a unit circle (and for small angles, the arc is essentially a straight line).

20

Your equation is not an ellipse, or even a family of ellipses. Ellipses are graphs given by a quadratic relation in $$Ax^2+Bxy+Cy^2+Dx+Ex+F=0.$$ Your equation, using Taylor Expansion (for $k\to\infty$), is $$\sum _{n=0}^k\frac{\left(-1\right)^n\left(x^{2n}+y^{2n}\right)}{\left(2n\right)!}=\sum ... 19 We will use a well-known^{[1]} formula for sum of arctangents:$$\arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right) \pmod \pi\tag1$$The exact equality holds for uv<1, for other values there is additional term — an integer multiple of \pi (it is easy to determine in each case). Using this formula, we can establish the following ... 17 It looks like an ellipse because the isocurves of any smooth surface look like an ellipse in the vicinity of an extremum ! Indeed, by Taylor's development in 2D,$$f(x,y)=f(a,b)+\frac{\partial f}{\partial x}(x-a)+\frac{\partial f}{\partial y}(y-b)\\ +\frac12\frac{\partial^2 f}{\partial x^2}(x-a)^2+\frac12\frac{\partial^2 f}{\partial x\partial ...

13

It seems to me that the best answer thus far is Simon S's. Others have hinted at the important property: $$\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1$$ Some have simply stated it's important with little reason as to why it's important (specifically in regards to your question about the derivative of $\sin(x)$ equaling the $\cos(x)$). Simon S's answer ...

12

I believe your confusion stems from a misperception of "sine" and "angle". We tend to think of the units we measure an angle in (degrees, or radians, or grads, or whatever) as being independent of the sine function. In truth, this is not so: the sine of an angle measured in degrees, and the sine of an angle measured in radians are two different functions. ...

12

It might help to forget, just briefly, that sine and cosine have anything to do with triangles. Instead, let's look at them in this light: Let the function $s(x)$ be the unique function with the property that $s''(x) = -s(x)$ such that $s(0) = 0$ and $s'(0) = 1$. So it's a function that, when differentiated twice, gives the same function with a change of ...

11

Andre already gave the answer, but let me explain the "generalities". The main problem is the following: After you reached the point $\sin(x+y) = \sqrt{2}/2$, you concluded that $x+y$ can take both values $\pi/4$ or $3\pi/4$. You interpreted this to mean BOTH $x+y = \pi/4$ and $x+y = 3\pi/4$ are solutions. (That is, the set of solutions is the ...

9

In your calculation of $x+y$ via $\sin(x+y)$, you wrote that $x+y=\frac{\pi}{4}$ is a possibility. It is not, since already $\sin y=\frac{9}{\sqrt{130}}\gt\frac{1}{\sqrt{2}}$, so $y\gt \frac{\pi}{4}$. Remark: In this case, computing $\cos(x+y)$ is the better strategy, since the cosine here unambiguously identifies $x+y$. Computing $\sin(x+y)$ also works, ...

9

Hint: consider $$x_{n}=a^n+b^n+c^n$$ $$x_{n+3}=(a+b+c)x_{n+2}-(ab+bc+ac)x_{n+1}+abcx_{n}$$ let $a=\tan^2{20}=\tan^2{\dfrac{\pi}{9}},b=\tan^2{40}=\tan^2{\dfrac{2\pi}{9}},c=\tan^2{80}=\tan^2{\dfrac{4\pi}{9}}$ and you can add $d=\tan^2{\dfrac{3\pi}{9}}=3$ It is easy to find $$a+b+c=30,ab+bc+ac=27, abc=3$$ because you can see here lab bhattacharjee full ...

7

They are not ellipses. To prove this, consider the closed curve lying above and directly to the right of the origin. Since the given equation is symmetric between $x$ and $y$, this curve is symmetric about the line $y=x$. Thus, if it is an ellipse, its minor axis must line on the line $y=x$. Now, it is easy to check that the points $$(\pi,\pi) \pm ... 7 Hint$$\int\dfrac{\cos{\theta}}{\sqrt{2-9\sin^2{\theta}}}d\theta=\int\dfrac{d\sin{\theta}}{\sqrt{2-9\sin^2{\theta}}}$$and$$\int\dfrac{1}{\sqrt{2-9t^2}}dt=\dfrac{1}{\sqrt{2}}\cdot\int\dfrac{1}{\sqrt{1-\left(\frac{3}{\sqrt{2}}t\right)^2}}dt=\dfrac{1}{3}\cdot\int\dfrac{1}{\sqrt{1-\left(\frac{3}{\sqrt{2}}t\right)^2}}d\left(\dfrac{3}{\sqrt{2}}t\right)$$6 The answer of @Vladimir Reshetnikov: is hard to improve on, so we'll try to answer some of the questions posed in the comments. If \phi=\frac{1+ \sqrt{5}}{2} is the positive root of x^2 - x-1=0 then \begin{eqnarray} \tan(2 \arctan(\phi^n)) = \frac{2\phi^n}{ 1- \phi^{2n}} = \begin{cases} -\frac{2}{L_{n}} \text{ if n odd} \\ -\frac{2}{F_{n}\sqrt{5}} ... 6 You are almost there. If the equation has rational roots, then \Delta must be a perfect square (why?). To show that \Delta is not a perfect square just observe that$$\Delta=96k^2-316k+11 \equiv 3 \pmod{4}$$5 This is an old question, but should be interesting to answer. If you want an aesthetic version, then the solution to,$$x^5 + x^4 - 12x^3 - 21x^2 + x + 5 = 0$$is given by,$$x = \frac{1}{5}\left(-1+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\right)$$and the z_i are the roots of the quartic,$$z^4 + 12679 z^3 + 78678031 z^2 + 362989005529 z + 31^{10} = ...

5

Simply: Solving for an angle using its sine value is ambiguous. (This is one reason that the Law of Cosines is preferred over the Law of Sines for determining angles from sides.) Here's an analogous situation: \begin{align}x^2 + 2 x = 35 \quad(1)\\ x^2 - 2 x = 15 \quad(2) \end{align} Adding $(1)+(2)$, we see that $2 x^2 = 50$, so that $x = \pm 5$. ...

4

Usually, yes, though I prefer Euler's identity. Pretty much every trig identity can be derived from $$e^{ix}=\cos(x)+i\sin(x).$$ However, it is useful to memorize some of the common ones because they will help you a lot in calculus and beyond to quickly identify when an expression can be simplified. I would start with memorizing the angle addition formulas. ...

4

Write the original expression as follows \begin{equation*} \frac{x\cos x-\sin x}{2x^{3}}=\frac{x\cos x-x+x-\sin x}{2x^{3}}=\frac{1}{2}% \left( \frac{\cos x-1}{x^{2}}+\frac{x-\sin x}{x^{3}}\right) . \end{equation*} Now use the standard limits \begin{equation*} \lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}}=-\frac{1}{2},\ \ \ \ \ \ and\ \ \ \ \ \ \ \ ...

4

The solution is nothing more than some computation and observing $2\cdot27 = 90 - 36.$ \begin{align*} \tan^2(27^\circ) + 2 \tan(27 ^ \circ) \tan (36^\circ) &= \tan^2(27^\circ) + 2 \tan(27 ^ \circ) \cot (54^\circ) \\[1ex] &= \tan^2(27^\circ) + 2 \, \frac{\tan(27 ^ \circ)}{ \tan(54^\circ)} \\[1ex] &= \tan^2(27^\circ) + 2 \, \frac{\tan(27 ^ ...

4

$1+\cot^2(-\theta)=1+(-\cot\theta)^2=1+\cot^2\theta=1+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}=\dfrac{1}{\sin^2\theta}=\csc^2\theta$

4

If $x=\pi/4$ we have: $$\frac{\sqrt{1+1}}{1}=\sqrt{2}$$ on the left, but $$\frac{1+ \frac{1}{\sqrt{2}}}{1/\sqrt{2}}=\sqrt{2}+1$$ on the right. These are not equivalent.

4

As stated, equality is false. Letting $x=0$, the left hand side is $0$ while the right hand side is $1$.

4

Square both equations. $$A^2\sin^2(x)+B^2\cos^2(x)+2AB\sin(x)\cos(x)=c^2,$$ $$B^2\sin^2(x)+A^2\cos^2(x)-2AB\sin(x)\cos(x)=d^2.$$ Now add then together. $$A^2(\sin^2(x)+\cos^2(x))+B^2(\sin^2(x)+\cos^2(x))=c^2+d^2,$$ $$A^2+B^2=c^2+d^2.$$ When solving these types of problems, it is important to remember trigonometric identities, like $\sin^2(x)+\cos^2(x)=1$ ...

4

It is correct to use the $\approx$ sign for rounded answers, and the $=$ sign for closed form answers. For example, if you are solving $\sin x=\cos x$, one solution would be $x=\frac{\pi}4\approx 0.866$. The use of an equal sign is probably because of a bad textbook, or that the writers used it because the rounded answer is very close to the real answer. ...

4

Hint: Note this indenty $$a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(a+c)$$ so $$(\sin{x}+\sin{(2x)})(\sin{(2x)}+\sin{(3x)})(\sin{x}+\sin{(3x)})=0$$ then $$\sin{x}\cos{x}(2\cos{x}+1)(4\cos^2{x}+2\cos{x}-1)=0$$ then you can solve it

3

The equal sign $=$ means that the LHS and the RHS indicate the same number so, if we write $\theta=\sin^{-1} \left(\frac {\sqrt{3}-1}{2} \right)$, this means that $\theta$ is exactly the real number indicated by the symbols in the RHS. The symbol of approximation $\approx$ means that the LHS and the RHS can differ by a quantity that is less than a value, ...

3

The direction is correct in general. But I recommend skipping the step in 2nd equation, $B=\frac A{\tan\theta}$, as when $\theta=0,\tan\theta=0$ and $B$ is not defined. It'd be better if you directly write $B=C\cos\theta$, in which $B$ is well-defined for all $\theta$. Btw, have you also considered the case beyond $0\le \theta \le \frac{\pi}2$?

3

It is well-known that the sum of the distances of the circumcenter from the sides of a triangle equals the sum of the circumradius and the inradius, hence: $$R\left(\cos A+\cos B+\cos C\right) = R + r \tag{1}$$ and since $R\geq 2r$ by Euler's theorem, we have: $$\cos A + \cos B+\cos C \leq \frac{3}{2} \tag{2}$$ with equality attained only by the ...

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