# Tag Info

0

Suppose you already know the identity $\sin^2 \phi + \cos^2 \phi = 1$. Then: \begin{align*} \frac{1}{\sqrt{1+\tan^2(\phi)}} &= \frac{1}{\sqrt{1+\frac{\sin^2 \phi}{\cos^2\phi}}}\end{align*}\\ =\frac{1}{\sqrt{\frac{\cos^2\phi}{\cos^2\phi}+\frac{\sin^2 \phi}{\cos^2\phi}}}\\ = \frac{1}{\sqrt{\frac{\cos^2\phi + \sin^2 \phi}{\cos^2\phi}}}\\ = \frac{1}{\... 1 I have no reference, but a derivation is that in the first quadrant is \frac{1}{\sqrt{1 + \tan^2(\phi)}} = \frac{1}{\sqrt{1 + \frac{\sin^2(\phi)}{\cos^2(\phi)}}} = \sqrt{\frac{\cos^2(\phi)}{\sin^2(\phi) + \cos^2(\phi)}} = \cos(\phi)$$and then$$ \frac{\tan(\phi)}{\sqrt{1 + \tan^2(\phi)}} = \tan(\phi) \cdot \frac{1}{\sqrt{1 + \tan^2(\phi)}} = \tan(\phi) \...

2

I suppose $p$ is an odd prime. Let the product be $S_p$ (it is positive). Using $$\ln\left|2\sin\frac{x}{2}\right|=-\sum_{m=1}^{\infty}\frac{\cos mx}{m}\qquad(x\notin2\pi\mathbb{Z}),$$ we see that $$\ln(2^{p-1}S_p)=-\sum_{m=1}^{\infty}\frac{l_p(m)}{m},\qquad l_p(m)=\sum_{n=1}^{p-1}\cos\frac{4mn^2\pi}{p}.$$ The article allows to find $l_p(m)=a_p(m)+b_p(m)$, ...

0

Let x=3°, y=60°, rewrite expression as $\{\frac{\sin(3x)}{\sin(x)}\} \div \{\sin(y-x)\sin(y+x)\}$. $$\frac{\sin(3x)}{\sin(x)} = \frac{3\sin(x)-4(\sin(x))^3}{\sin(x)} = 3 - 4(\sin(x))^2 = 1 + 2\cos(2x)$$ $$\sin(y-x)\sin(y+x) = \frac{1}{2} \{-\cos(2y) + \cos(2x)\} = \frac{1}{4}(1 + 2 \cos(2x))$$ Thus, expression simplified to 4

5

Let $\alpha = 60^\circ$ and $\theta = 3^\circ$. The numerator is \begin{align} \cos(81^\circ) = \sin(9^\circ) &= \sin(3\theta)\\ &= \sin\theta\cos(2\theta) + \cos\theta\sin(2\theta)\\ &= \sin\theta(\cos(2\theta) + 2\cos^2\theta)\\ &= \sin\theta(2\cos(2\theta) + 1)\end{align} while the denominator equals to \begin{align} \sin 3^\circ\... 3 Follow the steps below,\frac {\cos 81}{\sin3 \sin57 \sin63 }=\frac{2 \sin 9 / \sin 3 }{ \cos 6 - \cos 120 } =4\times \frac{1+2\cos 6}{1+2\cos 6}=4$$where, at various steps above, the followings are used,$$\cos(120)=-1/2\frac{\sin (3\times 3)}{\sin 3} =\frac{3\sin 3-4\sin^3 3 }{\sin 3} =1+2\cos 6 \cos 6 - \cos 120= \cos (...

1

For a directed angle $\theta$ with vertex at the origin and initial side on the positive $x$-axis, the point at which the terminal side of the angle $\theta$ intersects the unit circle $x^2 + y^2 = 1$ is defined to be $(\cos\theta, \sin\theta)$. If the terminal side of angle $\theta$ intersects the circle $x^2 + y^2 = r^2$ at the point $(x, y)$, then by ...

1

If you know an approximate of the solution, you can use the fixed-point method to find a less approximate (!) solution. Define $$\alpha_{n+1}=\sqrt[k+1]{{4.68-4.50\cos \alpha_{n}\over 1.23}\cdot \alpha_n^{k}}$$and find the limit. I tried $k=3$ and obtained $\approx 4.5166$. All you have to do is that using a scientific calculator press an approximate of ...

0

You can observe that $$\lvert\sin x\rvert=\sqrt{1-\cos^2x}=\sqrt{1-\frac{24^2}{25^2}}=\frac{7}{25}$$ Since $\pi/2<x<\pi$, we conclude that $\sin x=7/25$. Also $$\lvert\cos y\rvert=\sqrt{\frac{1}{1+\tan^2y}}=\frac{40}{41}$$ and from $\pi<y<3\pi/2$ we conclude that $\cos y=-40/41$. Thus $$\sin y=\tan y\cos y=-9/41$$

1

Your method is correct. Indeed, refer to the graphs: $\hspace{1cm}$ $$\cos x=-\frac{24}{25}=\frac{\overbrace{-24}^{adjacent}}{25}, x\in \left(\frac{\pi}{2},\pi\right); \quad \tan y=\frac{9}{40}=\frac{\overbrace{-9}^{front}}{\underbrace{-40}_{adjacent}},y\in \left(\pi,\frac{3\pi}{2}\right).$$ Also note: $$\cos x=-\cos(\pi -x)=-\frac{24}{25} \Rightarrow \cos ... 0 Yes, you are right. The law of sines helps, but for fife triangles.$$\frac{CD}{DB}=\frac{\frac{CD}{AD}}{\frac{DB}{AD}}=\frac{\frac{\sin(45^{\circ}-\alpha)}{\sin\alpha}}{\frac{\sin(45^{\circ}+\alpha)}{\sin\alpha}}=\frac{\sin(45^{\circ}-\alpha)}{\sin(45^{\circ}+\alpha)}.$$In another hand,$$\frac{CD}{DB}=\frac{\frac{CD}{ED}}{\frac{DB}{ED}}=\frac{\frac{\sin(...

1

If the question is referring to the first quadrant, where all trigonometric identities are positive, ... When you're told that the $\sin$ and $\cos$ of a quantity have equal sign, it doesn't necessarily mean they're both in the first quadrant (that happens only if you know both to be positive). What the statement means is that they're both either in the ...

0

Just do it. $\cos A = \frac 12$ So what is $\sin A$? Well $\cos^2 A + \sin^2 A = 1$ and so $(\frac 12)^2 + \sin^2 A = 1$ so $\sin^2 A= 1 - \frac 14$ so $\sin A =\pm \frac{\sqrt {3}}2$. But we are told $\sin A$ and $\cos A$ are the same sign so $\sin A =\frac {\sqrt 3}2$. So what is $\sin -A$? Well, we have a rule that $\sin (-A) = -\sin A$ (and ,yes, ...

0

Remember the negative angle identities: $$\sin(-A)=-\sin(A)$$ $$\cos(-A)=\cos(A)$$ To find the value of A, we have to solve the equation $\cos^{-1}(A)=\frac{1}{2}$ (Remember your 30-60-90 triangle) Assuming that this is over the interval $[0, 360]$, you can get either $60^{\circ}$ or $300^{\circ}$ (Remember your reference angle formulas) Because the ...

0

$\cos (\theta) =\frac {b}{a+c}$ $$\cos (\theta) =\frac{(Sin B)}{(Sin A\,+ SinC )}$$ $$\cos(\theta) = \frac{ Sin (B/2) }{Cos (( A -C)/2)}$$ $$\cos (\theta) =\frac { Cos( (A+C)/2) }{Cos (( A-C)/2)}$$ $$\cos (\theta) = \frac {1-(tanA/2)(tanC/2)}{1+(tanA/2)(tanB/2)}$$

1

Using $$\cos(\theta)=\frac{b}{a+c}$$ we get by squaring $$\frac{b^2}{(b+c)^2}=\cos^2(\theta)=1-\sin^2(\theta)$$

1

One way to solve the problem is to use the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$. \begin{align*} \sin^2\theta + \cos^2\theta & = 1\\ \cos^2\theta & = 1 - \sin^2\theta\\ |\cos\theta| & = \sqrt{1 - \sin^2\theta} \end{align*} Since $\theta$ is a second-quadrant angle, $\cos\theta < 0$. $$\cos\theta = -\sqrt{1 - \sin^2\theta}$$ ...

0

Since the two sine waves are out of phase, I don't think this will be possible, in general. I plotted an example: import numpy as np import matplotlib.pyplot as plt t = np.arange(0.00, 20.0, 0.1) A1 = np.sqrt(2) A2 = np.sqrt(3) B1 = np.sqrt(5) B2 = np.sqrt(7) C1 = np.sqrt(11) C2 = np.sqrt(13) data1 = A1*(np.sin(B1*t +C1)) data2 = A2*(np.sin(B2*t +C2)) ...

2

Suppose that $\tan z=i$. That means that $\sin z=i\cos z$. Then $$\sin^2 z+\cos^2 z=(i\cos z)^2+\cos^2z=(i^2+1)\cos^2 z=0.$$ But $$\sin^2 z+\cos^2 z=1.$$ Therefore there is no complex number with $\tan z=i$ (or with $\tan z=-i$).

3

I reckon this can be explained using ideas from algebraic geometry (of course, that specific theory only applies to polynomials, but you get the idea). When you prove some identity from some other condition, what you are really doing is to prove the ideal generated by the identity completely contains the ideal generated by the condition. So when you "use" ...

3

Why are you not comfortable about the structure of your first proof? With the little claim about the case when $\sin b=0$ mentioned in the comments under OP, it's OK. What you did is that you managed to reduce the problem to showing that the equality $$\frac{\cos^2a}{\cos^2b}=\frac 12$$ is true. Then you used the given hypothesis to prove this (more ...

0

If a triangle has the same proportions as one that fits a unit circle, it is called similar. Under these conditions, if you take the ratios of any $2$ sides, you will come up with the same $number(s)$ which mean the same angle(s) as though it fit the unit circle.

1

Starting from this line continue as follows (instead of dividing by $\cos ^2b$, express it as $1-\sin^2b$): $$(\sin^2a - \cos^2a) \cos^2b - 2\sin^2b\cos^2a = 0 \\ (\sin^2a - \cos^2a) (1-\sin^2b) - 2\sin^2b\cos^2a = 0\\ -\cos 2a-\sin^2a\sin^2b+\sin^2b\cos^2a-2\sin^2b\cos^2a=0 \\ -\cos2a-\sin^2a\sin^2b-\sin^2b\cos^2a=0\\ \cos2a+\sin^2b=0.$$

2

$\cos x$ is strictly decreasing and positive for $x\in [0,1].$ So $\cos a +\cos b$ cannot be a minimum if $a<c-b$ because then $\cos a +\cos b >\cos (c-b)+\cos b.$ For $a=c-b$ let $M=(a+b)/2=c/2$ and $D=(a-b)/2.$ We have $$\cos a=\cos (M+D)=\cos M \cos D-\sin M\sin D$$ and we have $$\cos b=\cos (M-D)=\cos M \cos D+\sin M \sin D.$$ Adding, we have (...

0

$\cos(a) + \cos(b) = \cos(a) + \cos(c_0 - a)$. The derivative with respect to $a$ is $-\sin(a)+\sin(c_0-a)$. For $0 < c_0 \le c < 1, 0 \le a \le c_0$, the derivative is clearly decreasing, from maximum $\sin(c_0)$ to minimum $-\sin(c_0)$. So the minimum must occur at either $a=0$ or $a=c_0$. The values at these two points are actually the same: $1+\cos(... 2 This is not an answer but it strengthen your guess. Suppose$a+b=2k\leq c$and let$a=k-\varepsilon$and$b=k+\varepsilon$then: $$f(a,b) :=\cos(a)+\cos(b)=2\cos(k)\cos(\varepsilon),$$ Now one can see that when$\varepsilon\to 0$,$f(a,b)$increases and for$\varepsilon\to k$,$f(a,b)$decreases. So probably the minimum occur when$\varepsilon=k$and$k\to c/...

1

$\cos(ix)=\cosh(x)$ and $\sin(ix)=i\sinh(x)$. Beyond that, most of the hyperbolic trig identities have a similar form compared to their "circle" trig counterparts. For instance, $$\sinh(x+y)=\sinh x\cosh y+\cosh x+\sinh y$$ $$\cosh^2x-\sinh^2=1$$ Also, as noted in the comments and on Wikipedia, there is a connection to the hyperbola $x^2-y^2=1$ ...

2

If you graph the functions (should be easy to do by hand. If you don't know how to graph the sine function, learn here), a solution is where the graphs intersect. Where c = 0: There is one solution between 0 and $\pi$. This proves option 3 and option 2 correct. Where c = 1: There are 3 solutions, two between $-\pi$ and 0 and one between 0 and $\pi$. This ...

2

You can use the fact that $f(x) = - \sin x$ is convex on $[0, \pi/2]$ (for example, because its second derivative is positive on $[0,\pi/2]$). The convexity property (the graph is below any arc) yields $$- \sin \left( (1-t) \times 0 + t \times \frac{\pi}{2} \right) \leq (1-t) \times f(0) + t \times f\left(\frac{\pi}{2} \right)$$ and $$- \sin \left( \frac{\... 1 Since the slope of \tan(x) can be big, 2x=\tan(x) is not stable over a large domain. \arctan(2x)=x-k\pi offers more stability; i.e. better convergence. Newton's Method gives$$ x_{n+1}=\frac{(\arctan(2x_n)+k\pi)\left(4x_n^2+1\right)-2x_n}{4x_n^2-1} $$When k=0, we get three attractors, depending on x_0. If -\frac12\lt x_0\lt\frac12, then x_n\... 2 Using$$\arctan x+\arctan y\stackrel{!}{=}\arctan\tan(\arctan x+\arctan y)\\=\arctan\frac{\tan\arctan x+\tan\arctan y}{1-\tan\arctan x\tan\arctan y}=\arctan\frac{x+y}{1-xy}$$(the conditions under which \stackrel{!}{=} is valid are left as an exercise, but they apply to the examples we'll consider), we have$$\arctan\frac{\pi}{18}+\arctan\frac{\pi}{3}=\...

4

The arctangent addition formula gives $$\arctan \frac{\pi}{18} + \arctan \frac{\pi}{3}=\arctan\frac{7\pi/18}{1-\pi^2/54}=\arctan\frac{21\pi}{54-\pi^2}$$ Thus the expression is of the form $\arctan z+\arctan-z$, and since $\arctan$ is an odd function the expression equals $0$.

1

There is a far simpler method which does not involve calculus and which can reliably solve many equations of this kind. It does need a calculator, but not a graphing one. Consider the graphs of $\tan x$ and $2x$ at the point of intersection near $\pi$. The tangent curves upwards while the linear function is a straight line. In these circumstances, any ...

1

Let $y=\frac{\pi}{2}-x$, to solve the complementary angle. $2x=\tan(x) → \frac{1}{\pi - 2y} = \tan(y)$ $$f(y) = \tan(y) - \frac{1}{\pi - 2y}$$ $$f'(y) = \sec(y)^2 - \frac{2}{(\pi-2y)^2}$$ Newton's method: $y_{i+1} = y_i - \frac{f(y_i)}{f'(y_i)}$, we got 0.5 → 0.407975 → 0.405237 → 0.405235 $$x=\frac{\pi}{2}-y ≈ 1.16556$$ If trig. functions not ...

5

Consider that you look for the zero of function $$f(x)=2x-\tan(x)$$ and forget the trivial solution $x=0$. On the other hand, if $x$ is a root $-x$ is a root too. Then, we need to focus on the positive solution. In the restricted interval, the derivative $f'(x)=2-\sec ^2(x)$ cancels when $x=\frac \pi 4$ which is a local maximum. So, to get an approximation, ...

1

The solution of the problem is, as you said, $\theta=\frac{\pi}{3}+2\pi n \text{ or } \theta=\frac{5\pi}{3}+2\pi n$ with $n\in \mathbb{Z}$ (you can add or subtract $2\pi$ and the cosine doesn't change). But the solution isn't $\theta=\frac{\pi}{3}+\pi k \text{ or } \theta=\frac{5\pi}{3}+\pi k$ with $k\in \mathbb{Z}$! For example, you can check what happens ...

2

$u=\sqrt {a\cos^{2}x+b\sin^{2}x} + \sqrt {b\cos^{2}x+a\sin^{2}x}$ Let $p = a \cos^2 x + b \sin^2 x$ $q = b \cos^2 x + a \sin^2 x$ and $u = \sqrt{p} + \sqrt{q}$ Then $u^2 = p + q + 2 \sqrt{pq}$ Now $p + q = a + b$ and $\displaystyle pq = \frac{(a+b)^2}{4} - \frac{(a-b)^2}{8} - \frac{(a-b)^2}{8} \cos 4x$ If $\cos 4x = -1$ then $u^2$ is maximum and ...

0

Hint. $$(a\cos^2x+b\sin^2 x)(b\cos^2x+a\sin^2x) = \frac 18\left((a+b)^2+4ab-(a-b)^2\cos(4x)\right)$$

1

Use Euler's formula to write it like: $$\frac{e^{ix}}{1-ce^{iy}} = \frac{\cos x+i\sin x}{1-c(\cos y+i\sin y)}$$ Multiply top and bottom by complex conjugate $1-c(\cos y-i\sin y)$. $$\frac{\cos x+i\sin x}{1-c(\cos y+i\sin y)} \cdot \frac{1-c(\cos y-i\sin y)}{1-c(\cos y-i\sin y)}$$ which simplifies to $$= \frac{\cos x-c(\cos x\cos y-\sin x\sin y)}{1+c^... 0 The function (x-1)^2 has a maximum value of 1 on the interval [0,2]. It follows that (x-1)^2-1 has a maximum of 0 there too. Now we know that for 0\le x\le2, we have that$$0\le \sin(\frac{πx}{2})\le 1.$$Thus,$$(x-1)^2-1-\sin(\frac{πx}{2})=x^2-2x-\sin(\frac{πx}{2})\le 0$$on this interval. 1 Hilbert modular forms can be used to construct a curious number field, namely a non-solvable finite Galois extension of \mathbb{Q} that is ramified at p=2 only (well, this extension is also ramified at infinity if you count it). I guess it would be difficult to find this number field from scratch. See https://arxiv.org/abs/0811.4379, also https://... 1 Automorphic forms, a generalization of modular forms, can be used to prove a conjecture of Deligne stating that for a normal variety X over a finite field F_q, a prime number l relatively prime to q, an irreducible lisse l-adic sheaf \sigma on X whose determinant has finite order and a closed point x\in |X|, the roots of the polynomial \... 3 Since x \in [0,\pi] the condition -1\leq \tan \, x \leq 1 is same as 0 \leq x \leq \frac {\pi} 4 or \frac {3\pi} 4 \leq x \leq \pi. For 0 \leq x \leq \frac {\pi} 4 the inequality \sin (2x) \geq 0.5 is equivalent to 2x \geq \frac {\pi} 6. Can you handle the case \frac {3\pi} 4 \leq x \leq \pi? 0 Let Z is a coplex number, Z=r(cos x +i sin x), where i=√-1 Whe have from de moivre's theorem for any n Z^n = cos nx +i sin nx , Hence (cos nx+i sin nx)^3 = cos 3x +i sin nx By expanding and equting we get that Sin3x =3cos^2 x*sinx - sin^3 x 0 If the coordinates of \text{A} and \text{B} are \ x_A, y_A\  and \ x_B, y_B\  respectively, then those of \text{B2} will be \ \frac{x_A\cot b + x_B\tan a }{\cot b+ \tan a}, \frac{y_A\cot b + y_B\tan a }{\cot b+ \tan a}\ . To see this, drop a perpendicular from \text{B2} to \text{BC} to intersect it at \text{D}. Then \ \lvert \text{BD}\... 1 Let \dfrac{\pi-A}4=x etc. \implies4(x+y+z)=3\pi-\pi\iff x+y+z=\dfrac\pi2 Now as \tan x,\tan y,\tan z are real,$$(\tan x-\tan y)^2+(\tan y-\tan z)^2+(\tan z-\tan x)^2\ge0\implies\tan^2x+\tan^2y+\tan^2z\ge\tan x\tan y+\tan y\tan z+\tan z\tan x$$Finally$$\tan(x+y)=\tan\left(\dfrac\pi2-z\right)\iff\dfrac{\tan x+\tan y}{1-\tan x\tan y}=\...

2

$$(\cos x+i\sin x)^3=\cos^3x+3i\cos^2x\sin x+3i^2\cos x\sin^2x+i^3\sin^3x$$ $$=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x$$ $$=(\cos^3x-3\cos x\sin^2x)+i(3\cos^2x\sin x-\sin^3x)$$ Since this is $\cos 3x +i\sin3x$, we can conclude that $\cos3x=\cos^3x-3\cos x\sin^2x$ and $\sin3x=3\cos^2x\sin x-\sin^3x$. You can simplify with the Pythagorean identities ...

-1

Expand the right hand side, i.e. the term $( ... )^3$; Gather the imaginary part in the right hand side Identify the imaginary parts.

1

$$\left(\tan^2\frac{\pi-x}{4}\right)''=\frac{2-\sin\frac{x}{2}}{8\cos^4\frac{\pi-x}{4}}>0.$$ Thus, by Jensen $$\sum_{cyc}\tan^2\frac{\pi-\alpha}{4}\geq3\tan^2\frac{\pi-\frac{\alpha+\beta+\gamma}{3}}{4}=1.$$

0

I am assuming that your objective is to compute the length of $AB_2$ Let me call it $x$ and also let $\angle ACB_2 = \alpha$ In trigonometry, it is a standard convention to denote the side $AC$ by $b$, the side $BC$ by $a$ and the side $AB$ by $c$ Now by the Law of Sines, we can write \$\displaystyle \frac{x}{\sin \alpha} = \frac{b}{\sin (\pi - A - \...

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