# Tag Info

28

Use the fact that $$\tan{\left (\frac{\pi}{2}-x\right)} = \frac{1}{\tan{x}}$$ i.e., $$\frac1{1+\tan^{\alpha}{x}} = 1-\frac{\tan^{\alpha}{x}}{1+\tan^{\alpha}{x}} = 1-\frac1{1+\frac1{\tan^{\alpha}{x}}} = 1-\frac1{1+\tan^{\alpha}{\left (\frac{\pi}{2}-x\right)}}$$ Therefore, if the sought-after integral is $I$, then $$I = \frac{\pi}{2}-I$$ and...

11

Since all methods are accepted, take the complex exponential defined as its series and consider the complex definitions of the trigonometric functions: $$\cos (z)=\dfrac{e^{iz}+e^{-iz}}{2}\, \land \, \sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}, \text{ for all }z\in \mathbb C.$$ Take $\theta \in\mathbb R$. The following holds: \begin{align} (\cos(\theta))^2+(\sin ... 9 Rewrite as\int_{-\pi/2}^{\pi/2} \frac{dx}{1+e^{b x}} \frac1{1+\cot^{2012}{x}}$$where b=\log{\alpha}. This integral is equal to$$\underbrace{\int_{-\pi/2}^0 \frac{dx}{1+e^{b x}} \frac1{1+\cot^{2012}{x}}}_{x \mapsto -x} + \int_0^{\pi/2} \frac{dx}{1+e^{b x}} \frac1{1+\cot^{2012}{x}}$$or$$\int_0^{\pi/2} \frac{dx}{1+\cot^{2012}{x}}\underbrace{\left ...

8

Try the geometric definition of cosine: $|\cos \theta|=\frac{|\text{adj}|}{|\text{hyp}|}$. Since the hypotenuse is the longest side in a right triangle, we find that the ratio is at a maximum when $|$adj$|=|$hyp$|$. We can also find an instance of this happening, that is when $\theta=0$, or there is no opposite side. Therefore $$|\cos \theta|\le ... 6 Well it comes directly from the Pythagorean theorem. We know that in a right triangle, cos {\theta}=\frac{h}{r} and sin{\theta}=\frac{v}{r}, h is short for horizontal and v for vertical, r is the hypotenuse. Now, from the Pyth. theorem$$r^2=v^2+h^2=r^2 sin^2{\theta}+r^2 cos^2{\theta} \Leftrightarrow cos^2{\theta}+sin^2{\theta}=1$$By the way, ... 6 Consider a right-angled triangle ABC where <BAC = \theta, By Pythagorean theorem,$$\Large AC^2+BC^2=AB^2$$Dividing by AB^2,$$\Rightarrow \large\frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\Rightarrow \large(\frac{\text{opposite}}{\text{hypotenuse}})^2+(\frac{\text{adjacent}}{\text{hypotenuse}})^2 = \frac{AB^2}{AB^2}$$... 5 These are both immediate using two subsitutions. Everything is even, so split all of them at 0.$$\begin{aligned} t=\sqrt{1-x^2}:\quad\int_0^1 \frac{dx}{\sqrt{1-x^2}}= \int_0^1 2\sqrt{1-t^2}\,dt\end{aligned}$$And:$$\begin{aligned} t=\frac{x}{\sqrt{1-x^2}}:\quad\int_0^{1} \frac{dx}{\sqrt{1-x^2}}= \int_0^{\infty}\frac{dt}{1+t^2}\end{aligned}$$5 The answer given by Gordon is very good for computing the integral. But it does not provide much why it works. The following figure might help with that This is the integral for \alpha=\sqrt{2}. It looks as if the area under the function is exactly half of the dashed rectangle... A good guess is therefore$$ \int_0^{\pi/2} \frac{\mathrm{d}x}{1 + ...

5

Assuming the First Pythagorean Trigonometric Identity, $$\sin^2\theta + \cos^2\theta = 1$$ Dividing by $\cos^2\theta$, $$\Rightarrow \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$$ $$\Rightarrow \left(\frac{\sin\theta}{\cos\theta}\right)^2 + \left(\frac{\cos\theta}{\cos\theta}\right)^2 = ... 5 Here is an easier approach. For \alpha =0  or \pi/2, the left hand side makes no sense. Hence, we can take \alpha \in (0,\pi/2). We then have \tan(\alpha) to be positive. Now by AM-GM, for a,b \in \mathbb{R}^+, we have$$a+b \geq 2\sqrt{ab}$$Hence, we have$$\tan(\alpha) + \cot(\alpha) = \tan(\alpha) + \dfrac1{\tan(\alpha)} \geq 2 \cdot ...

5

If you know about equivalents, then you can say that $\tan(y)$ is equivalent to $y$ in zero and $\sin(y)$ is equivalent to $y$ in $0$. This gives that the limit you're looking is the same as the limit, when $x$ tends to $0$, of $\frac{8x}{3x}$, which is of course $\frac{8}{3}$. Here is an alternative proof if you don't know about equivalents: \begin{align} ...

4

I usually like to reduce these to problems involving complex exponents. Note that $$\sin\alpha\cos\beta + \sin\beta\cos\alpha = \frac{e^{i\alpha} - e^{-i\alpha}}{2i}\frac{e^{i\beta} + e^{-i\beta}}{2} + \frac{e^{i\beta} - e^{-i\beta}}{2i}\frac{e^{i\alpha} + e^{-i\alpha}}{2}$$ and simplify this expression.

4

To find the RHS (assuming that we may forget the result) we write: $$\sin(\alpha+\beta)=\Im\frac{1}{2}(e^{i(\alpha+\beta)}-e^{-i(\alpha+\beta)})\\=\Im\frac{1}{2}((\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta)-(\cos\alpha-i\sin\alpha)(\cos\beta-i\sin\beta))$$ and we develop and we take the imaginary part we find the desired formula.

4

Using the scalar product : let $A$ and $B$ be the points on the unit circle at arc length $a$ and $b$. Then $A(\cos(a);\sin(a))$ and $B(\cos(b);\sin(b))$, and $\widehat{BOA}=a-b$. Therefore : \begin{aligned} \overrightarrow{OB}\cdot\overrightarrow{OA} &= OA\times OB\times \cos(\widehat{\overrightarrow{OB};\overrightarrow{OA}})\\ &= 1\times 1\times ...

4

The above method is really verifying and always quick. Another method to arrive at the answer is by rationalising denominator (mainly when the answer [or RHS] is not known or one is asked to work out only from LHS to RHS): $$\frac{\sin x - \cos x + 1 }{\sin x + \cos x - 1 }\cdot \frac{\sin x + \cos x + 1}{\sin x + \cos x + 1}$$ $$\frac{ (\sin x + 1)^2 - ... 4 Here is an alternative using exponential forms:$$ \begin{align*} 1+\tan^2 \theta &=1+\left( \frac{e^{i \theta}-e^{-i \theta}}{i\left( e^{i \theta}+e^{-i\theta} \right)} \right)^2 \\ &=1-\frac{\left( e^{i \theta}-e^{-i \theta} \right)^2}{\left( e^{i \theta}+e^{-i\theta} \right)^2} \\ &=\frac{\left( e^{i \theta}+e^{-i\theta} \right)^2-\left( e^{i ...

4

I will attempt to simplify the integral. It might take me a while and I might edit this answer a few times, so this is not its final form yet. Let $D$ be the differential operator. We know $\displaystyle\int \dfrac{dx}{\sqrt{x^2-1}}=\ln(2(x+\sqrt{x^2-1}))$. So by using integration by parts we reduce the problem to solving ...

4

If you choose to define sine and cosine by trigonometric rations, then JohnK's answer answers your question. There are other ways of answering your question that go with the different definitions of sine and cosine. Here are a few: $(1)$, $\sin(x)$ is the solution to the differential equation $y''=-y$, $y(0)=0$, $y'(0)=1$, and $cos(x)$ is its derivative. ...

3

You should have found that $|\sin 2t\sin^2t|\le \left(\frac{\sqrt3}2\right)^3$. Thus the result follows from $$\left(\prod_{k=0}^n \sin(2^{k}t)\right)^3=\prod_{k=0}^n \sin^3(2^{k}t)=\underbrace{\sin t}_{|\cdot|\le 1}\cdot \prod_{k=0}^{n-1} \underbrace{\sin(2^{k+1}t)\sin^2(2^kt)}_{|\cdot|\le\left(\frac{\sqrt3}2\right)^3}\cdot ... 3 Assuming the First Pythagorean Trigonometric Identity,$$\sin^2\theta + \cos^2\theta = 1$$Dividing by \sin^2\theta,$$\Rightarrow \frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{sin^2\theta}\Rightarrow (\frac{\sin\theta}{\sin\theta})^2 + (\frac{\cos\theta}{\sin\theta})^2 = (\frac{1}{sin\theta})^2$$Since ... 3 no need to be too complicated here: both sin \;x and cos \;x are perodic, each with period 2\pi so sin \; 2x is periodic, with period \frac{2\pi}{2} = \pi and cos \; 8x is periodic, with period \frac{2\pi}{8} = \frac{\pi}4 can you find the least common multiple of \pi and \frac{\pi}4? do you see why that must be the period of ... 3 Let \#(a) be the number of real solutions to the equation \sin x = ax. For a \lesssim 1 we have \#(a) = 3, and this remains constant as a decreases until a = a_0, when the line y = ax is tangent to the curve y = \sin x. This point of tangency occurs just to the left of x = 2\pi + \tfrac{\pi}{2}. Here \#(a_0) = 5, and then for a ... 3 The differential equation$$f''=-f$$has a unique solution for given initial conditions f(0)=x_0, f'(0)=y_0. To show uniqueness assume that g is another solution for the same initial values. Then h:=f-g also satisfies h''=-h, and h(0)=h'(0)=0. But then$$\frac\partial{\partial t}(h(t)^2+h'(t)^2)=2h(t)h'(t)+2h'(t)h''(t)=2h(t)h'(t)-2h'(t)h(t)=0, $$... 3 I like to see it using Euler's formula e^{i\theta} = \cos{\theta} + i \sin{\theta}. We have$$ \begin{eqnarray} e^{i(\alpha+\beta)} = \cos(\alpha+\beta) + i \sin(\alpha+\beta) \end{eqnarray} $$but also$$ \begin{eqnarray} e^{i(\alpha+\beta)} &=& e^{i\alpha}\cdot e^{i\beta} \\\\ &=& [\cos(\alpha) + i \sin(\alpha)]\cdot [\cos(\beta) ...

3

What you should know to solve this problem is that, $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$ In your case you get, $$\frac{\tan(\alpha-\beta)+\tan\beta}{1-\tan(\alpha-\beta)\tan\beta}=\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$$ Now you know what $\sin\alpha$ is, but can you also try to find what is $\cos \alpha$ ?

2

If $A,B$ are the rest two acute angles, we immediately have $A+B=(180-105)^\circ=75^\circ$ On simplification of the given expression, $$\frac3{\cos x}=3(\cos x+\sin x)-n(\cos x-\sin x)$$ $$\implies \frac3{\cos x}=\sin x(3+n)+\cos x(3-n)\ \ \ \ (1)$$ $$\implies 3=(3+n)\sin x\cos x+(3-n)\cos^2x \ \ \ \ (2)$$ Multiplying either sides by ...

2

Consider the vectors $$\vec{u}=(\sin \alpha,\cos \alpha)= \left(\cos \left(\frac{\pi}{2}-\alpha\right),\sin\left(\frac{\pi}{2}-\alpha\right)\right)$$ and $$\vec{v}=(\cos \beta,\sin \beta)$$ in $\mathbb R^2$. The dot product $\vec{u} \cdot \vec{v}$ can be calculated in two ways: $\vec{u} \cdot \vec{v}=\|\vec{u}\| \|\vec{v}\| \cos \theta$ , where ...

2

If you don't want to use geometry, you can do it with complex powers of $e$: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}\\ \cos(x)=\frac{e^{ix}+e^{-ix}}{2}\\ \sin(x+y)=\frac{e^{i(x+y)}-e^{-i(x+y)}}{2i}$$ You can now also express the right hand side of your equality in terms of powers of $e$, and this wil give the same result.

2

Let your point $(x, y)$ be $A$. You gotta find the point a distance $n$ away at a given angle. Let $B(x_1,y_1)$ be your required point.. The easiest way to solve this, is to break it into components along the x and y axes. By simple trigonometry: $x_1 = x + n\cos\theta$ $y_1 = y + n\sin\theta$ Where $\theta$ is the angle given to you.

2

You've got it almost correct, but there's a mistake: You should conclude that $$\sin{\alpha} \cos{\alpha} \le \frac 1 2$$ not the other way around. Here's a very different approach: The function $f(x) = \tan{x} + \cot{x}$ blows up at $0$ and $\frac{\pi}{2}$, but is differentiable in between these points. We then have 0 = f'(x) = \sec^2{x} - ...

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