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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 $$4x=t\iff4\,\mathrm dx=\,\mathrm dt$$ $$I=\int\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm dx=\frac14\int\frac{\sin(2t)}{9+\sin^4(t)}\,\mathrm dt =\frac14\int\frac{2\sin t\cos t}{9+\sin^4(t)}\,\mathrm dt$$ 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, $$\int \csc^6 2x ~dx = \frac{1}{2} \int \csc^6 \theta ~d\theta$$ $$ = \frac{1}{2} \int(1+\cot^2\theta)^2\csc^2\theta ~ d\theta$$ If $u = \cot \theta$, 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 $2x=\sec\theta$ $$\implies4x^2-1=\tan^2\theta$$


3

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.



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