# Tag Info

13

I tried writing $4x$ as $3x+x$ and $2x$ as $3x-x$ to get:$$\sin(3x+x)+\sqrt{3}\sin(3x)+\sin(3x-x)=0$$$$\therefore\sin(3x)\cos(x)+\cos(3x)\sin(x)+\sqrt{3}\sin(3x)+\sin(3x)\cos(x)-\cos(3x)\sin(x)=0$$$$\therefore2\sin(3x)\cos(x)+\sqrt{3}\sin(3x)=0$$$$\therefore\sin(3x)(2\cos(x)+\sqrt{3})=0$$Hopefully you can solve from here...

10

Let $\theta = \arcsin (\tan \alpha) \to \sin \theta = \tan \alpha \to 1 - 2\tan^2\alpha = 1 - 2\sin^2\theta = \cos 2\theta \to \arccos \left(1-2\tan^2\alpha\right) = \arccos (\cos 2\theta) = 2\theta \to L = \dfrac{2\theta}{2\theta} = 1$

8

$1 - \dfrac{\sin^2(2x)}{2} = \dfrac{1+\cos^2(2x)}{2}$, and $\sin x\cos x = \dfrac{\sin (2x)}{2} \Rightarrow \displaystyle \int \dfrac{\sin x\cos x}{\cos^4x+\sin^4x}dx = \displaystyle \int -\dfrac{1}{2}\dfrac{d(\cos(2x))}{1+\cos^2(2x)}dx = -\dfrac{1}{2}\arctan(\cos (2x)) + C$

7

OK I am going to write something else. Using sum to product formula, \begin{align*} \sin 2x + \sin x &= 0 \\ 2\sin\frac{3x}2\cos\frac{x}{2} &= 0\\ \sin\frac{3x}2 &= 0&\text{or}& &\cos\frac{x}{2} &= 0\\ \end{align*} Since $x\in[0,2\pi)$, $\frac{3x}2\in[0,3\pi)$ and $\frac x2\in[0,\pi)$. So the first case gives $$\frac{3x}2 = 0 ... 6 Rewriting the equation using the double-angle identity for \sin gives$$2 \sin x \cos x + \sin x = 0,$$and factoring gives$$(2 \cos x + 1) \sin x = 0.$$This holds iff$$2 \cos x + 1 = 0 \qquad \text{or} \qquad \sin x = 0.$$6 let radius = r, then in triangle ACO, using Pythagoras:$$\begin{align} AO^2 &= AC^2+CO^2\\ \\ r^2 &= 12^2+(r-5)^2\\ \\ 144+25-10r &= 0\\ \\ r &= 16.9\\ \end{align}$$In triangle ACO,$$\begin{align} \sin\theta &= \dfrac{12}{r}\\ \\ \sin\theta &= \dfrac{12}{16.9}\\ \\ \theta &= 45.2^{\circ} (1dp)\\ \end{align}$$6 For the error, see the comments and other answers which address it already. If you are familiar with Taylor series, you can compute your limit as follows, observing that$$ \sin y \operatorname*{=}_{y\to 0} y - \frac{y^3}{6} + \frac{y^5}{120} +o(y^5)\ . \begin{align} \sin (x+\frac{x^3}{6}) &\operatorname*{=}_{x\to 0} x+\frac{x^3}{6} - ...

6

The $\tan$ function is defined on $\Bbb R\setminus \left\{\frac{\pi}2+k\pi,\; k\in\Bbb Z\right\}$ and it's continuous on this set. You can't say that this function is discontinuous on $\frac{\pi}2+k\pi$ since it isn't even defined on these points.

5

This depends on what you're allowed to use. One thing we can say is $$\sin(x+\delta)-\sin(x)=\sin(x)\cos(\delta)+\sin(\delta)\cos(x)-\sin(x) \\ = \sin(x) (\cos(\delta)-1) + \sin(\delta) \cos(x)$$ Now you can control both terms by making $\delta$ small enough. In particular, you can prove, using only trigonometry, that both terms are no larger in magnitude ...

5

Well.. I learned this proof for the inequality. Let $x,y\in\mathbb{R}^k$. Observe that $|x+y|^2=(x+y)(x+y)=|x|^2+|y|^2+2xy\le |x|^2+|y|^2+2|xy| \le|x|^2+|y|^2+2|x||y|=(|x|+|y|)^2$ In the last inequality, we have used the Cauchy/Schwarz Inequality ($|xy|\le|x||y|$). For $k>1$ multiplication is exchanged by vector product.

5

Why not add $\sin 4x$ and $\sin 2x$? You'll get $2\sin 3x\cdot \cos x +\sqrt{3}\sin 3x = 0$...

4

You know $x, y, a, b$ and $r$. Then $$\cos(t) = \frac{x-a}{r} = c_{1} \quad \text{and} \quad \sin(t) = \frac{y-b}{r} = c_{2}$$ Note that $c_{1}$ and $c_{2}$ are just two numbers that you compute based on what you know. How many solutions does each of the above trigonometric equations have in $t \in [0, 2\pi)$? Of course, in the end, the right $t$ should ...

4

The slope of the line is $3$ . So, $\frac{dy}{dx}=3$ 1+\frac{2}{x^3}=3$$\implies x=1 and y=0 Equation of the line is (y-0)=3(x-1) \implies y=3x-3 4 Please refer to the trigonometric inverse functions, especially arcsine and arccosine. Given that \cos t = c_1 and \sin t = c_2, Fundamentally, for a certain range, \displaystyle t = \arccos(c_1) = \arcsin(c_2) Since \sin(x) and \cos(x) are many-to-one functions, there will correspond multiple values of x that yield a certain \sin(x) or ... 4 First realize that \sin 2x = 2 \sin x \cos x . Then, we have$$ 2 \sin x \cos x + \sin x = 0 \iff \left( 2 \cos x + 1 \right) \cdot \sin x = 0, $$which is ture iff \cos x = - \frac {1}{2} or \sin x = 0 . Can you solve each equation and find the family of solutions from here? 4 Alternatively, one could rewrite the equation in complex form:$$\frac{1}{2i}(e^{2 i x} - e^{-2 i x}) + \frac{1}{2i}(e^{i x} - e^{-i x}) = 0.$$Multiplying through by the (never-zero) quantity 2 i e^{2 i x} gives$$e^{4ix} + e^{3ix} - e^{ix} - 1 = 0. \qquad (\ast)$$The left-hand side is just the polynomial$$z^4 + z^3 - z - 1 = (z + 1)(z^3 - 1)$$... 4$$\sin2x=-\sin x=\sin(-x)\implies2x=n\pi+(-1)^n(-x)$$where n is any integer If n is even, =2m(say)$$2x=2m\pi-x\iff x=\frac{2m\pi}3$$If n is odd, =2m+1(say)$$2x=(2m+1)\pi+x\iff x=(2m+1)\pi$$4 HINT: Let t=x^2, dt=2x\,dx and the rest should be simple. So: yes, there is a mistake in your solution. Your answer cannot be identical with the correct one, because the former function is periodic and the latter not. 4 Hint$$f(x)=\sqrt 3 \sec(2x+\frac{\pi}{2})-2=\frac{\sqrt 3}{\cos(2x+\frac{\pi}{2})}-2=\frac{-\sqrt 3}{\sin(2x)}-2$$So, if f(x)=0,$$\sin(2x)=-\frac{\sqrt 3}{2}$$I am sure that you can take from here but, just as Przemysław Scherwentke answered, I am not sure at all that the choices given in the book are correct. 4 Let$$I=\int\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm dx$$Substitute$$4x=t\iff4\,\mathrm dx=\,\mathrm dtI=\int\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm dx=\frac14\int\frac{\sin(2t)}{9+\sin^4(t)}\,\mathrm dt =\frac14\int\frac{2\sin t\cos t}{9+\sin^4(t)}\,\mathrm dt$$Again substitute$$\sin^2 t= u\iff 2\sin t\cos t \,\mathrm dt=\,\mathrm du$$... 4 In figure 1, G is the intersection of the circles C_1: x^2 + y^2 = (2s)^2 and C_2: (x – s)^2 + (y – s)^2 = s^2. For some k, C_3 : C_1 + kC_2 = 0 is a family of circles passing through G (and G’). For a suitable k, we can generate the circle C_3 : (x – 2s)^2 + (y – 2s)^2 = (\sqrt (2)s)^2, which has center at C(2s, 2s) and radius = \sqrt ... 4 Hint: Recall that \csc^2 \theta = 1 + \cot^2 \theta. Let us perform the substitution \theta = 2x. Then,$$\int \csc^6 2x ~dx = \frac{1}{2} \int \csc^6 \theta ~d\theta = \frac{1}{2} \int(1+\cot^2\theta)^2\csc^2\theta ~ d\thetaIf u = \cot \theta, then du = \cdots? 4 \begin{align} \int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}\mathrm dx&=\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\left(1-\sin^2x\right)^2}\mathrm dx\\[7pt] &=\int_0^{\pi/2}\frac{\sin x\cos x}{2\sin^4x-2\sin^2x+1}\mathrm dx\\[7pt] &=\frac14\int_0^1\frac{\mathrm dt}{t^2-t+\frac12}\qquad\color{blue}{\implies}\qquad t=\sin^2x\\[7pt] ... 4 Here is one line proof\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx=\int_0^{\pi/2}\frac{\tan(x) (\tan(x))'}{\tan^4(x)+1}\ dx=\int_0^{\infty} \frac{x}{x^4+1}\ dx=\left[\frac{\arctan(x^2)}{2}\right]_0^{\infty}=\frac{\pi}{4}$$Q.E.D. 3 HINT: Using Trigonometric substitutions, set 2x=\sec\theta$$\implies4x^2-1=\tan^2\theta3 Hint: \sin x + \cos x = 0 if and only if \sin x = -\cos x if and only if \tan x = -1. 3 Call the matrix M. Then by using @r9m's suggestion, we interchange the two middle columns (this switches the sign of the determinant) and apply two row replacements (this doesn't change the determinant) in order to obtain a zero lower left submatrix: \begin{align*} |M| &= -\begin{vmatrix} \sin x & \cos x & \sin 2x & \cos 2x \\ \cos x & ... 3\sin{(2x)}+\sin{(x)}=2\sin{(x)}\cos{(x)}+\sin{(x)}=\sin{(x)}(2\cos{(x)}+1)=0\rightarrow\sin{(x)}=0\rightarrow x=0\space and\space x=\pi$$or$$\cos{(x)}=\frac{-1}{2}\rightarrow x=\frac{2\pi}{3}\space and\space x=\frac{4\pi}{3}$$3 Your mistake is in this line$$\lim_{x\to0}\dfrac{\left[{\left(x+\tfrac{x^3}6\right)\sin\left(x+\tfrac{x^3}6\right)}\left/\right.{\left(x+\tfrac{x^3}6\right)}\right]-x}{x^5}=\lim_{x\to0}\dfrac{x+\tfrac16x^3-x}{x^5}.$$You can't evaluate the limit using \lim\limits_{u\to0}\frac{\sin u}u=1 directly since you're using the two laws ... 3 Hint: Write it as$$\lim_{x\to0}\dfrac{\sin x-x}{x\sin x},\$ and apply L'Hôpital's rule twice.

Only top voted, non community-wiki answers of a minimum length are eligible