# Tag Info

## Hot answers tagged trigonometry

33

You seem to be asking why we name them both, rather than why they exist, since the very relationships you've written shows that if one exists, the other does, too. Essentially, all mathematical notation and names, except for a very small subset, are for convenience/clarity/human communication. It is not required that we name any function consistently, but ...

21

$$\sin^2 \alpha + \cos^2 \alpha =1,$$ $$\sin^2 \alpha + \sin^2 \left( \alpha + \frac{\pi}{2} \right) =1.$$ Which one do you prefer? Post scriptum: of course this is not a deep answer, but I think that sometimes mathematicians prefer elegance to logical "economy".

13

Using the Taylor series expansion, we can express $\sin x$ and $\cos x$ as $$\displaystyle\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$$ $$\displaystyle\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$$ Note that when $x \ll 1$, then $x^2 \ll x$ and $x^3 \ll x^2$ and so on. Any higher powers of $x$ will be very small. Then the above ...

13

Define $I(a)=\int_0^{\infty} e^{-(x+1)a}\frac{\sin(x)}{x+1}dx$. Then $\displaystyle I'(a)=-\int_0^{\infty} e^{-(x+1)a}\sin(x)dx=-\frac{e^{-a}}{a^2+1}$, and since $\lim_{a\to \infty} I(a)=0$, we have $\ \displaystyle I(0)=\int_0^{\infty} \frac{\sin(x)}{x+1}=\int_0^{\infty}\frac{e^{-a}}{a^2+1}da=\int_0^1\frac{dx}{1+\ln^2 x}.$

12

Hint. You may observe that $$\frac1{1+\ln^2 x}=-\Im \frac1{i-\ln x}=-\Im \int_0^{\infty}e^{-(i-\ln x)t}dt,\quad x \in (0,1),$$ giving \begin{align} \int_0^1 \frac{1}{1+\ln^2 x}\,dx&=-\Im \int_0^1\!\!\int_0^{\infty}e^{-(i-\ln x)t}dt\:dx\\\\ &=-\Im \int_0^{\infty}\!\!\left(\int_0^1x^t dx\right)e^{-it}dt\\\\ &=-\Im ... 9 Yes, \cos(x)^x is not hard to "explore". I don't know if it's been the subject of any published work, other than OEIS sequence A215347. It satisfies, for example, the differential equation y y'' = (y')^2 - (2 \tan(x) + x \sec^2(x)) y^2$$and it has the Maclaurin series$$ y = 1 - \dfrac{x^3}{2} - \dfrac{x^5}{12} + \dfrac{x^6}{8} - \dfrac{x^7}{45} + ...

8

Rewrite this as $$\sin^2 \theta - \cos^2 \theta = \sin^4 \theta - \cos^4 \theta$$ and then factor the right-hand side as a difference of two squares.

8

Let $\displaystyle \tan^{-1}\left(\frac{1}{2}\right) = \theta\;,$ Then $\displaystyle \frac{1}{2} = \tan \theta$ So $\displaystyle \cos\left[\tan^{-1}\left(\frac{1}{2}\right)\right] = \cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{1+\tan^2 \theta}}= \frac{2}{\sqrt{5}}$

7

For $a$ and $b$ positive numbers we have \begin{align*} \lim_{x\to 0^+}\frac{\sin \sqrt{ax}}{\sin \sqrt{bx}}=\lim_{x\to 0^+}\frac{\frac{\sqrt{a}\sin \sqrt{ax}}{\sqrt{ax}}}{\frac{\sqrt{b}\sin \sqrt{bx}}{\sqrt{bx}}} &=\frac{\sqrt{a}}{\sqrt{b}}\lim_{x\to 0^+}\frac{\frac{\sin \sqrt{ax}}{\sqrt{ax}}}{\frac{\sin ...

7

\begin{align}\frac{\sin 80^\circ \sin 20^\circ\sin 100^\circ }{\sin 50^\circ}&=\frac{\sin 80^\circ \sin 20^\circ\cdot 2\sin 50^\circ \cos 50^\circ }{\sin 50^\circ}\\&=2\sin 80^\circ\sin 20^\circ \cos 50^\circ\\&=2\left(-\frac 12(\cos 100^\circ-\cos 60^\circ)\right)\cos 50^\circ\\&=(\cos 60^\circ-\cos 100^\circ)\cos 50^\circ\\&=\frac ... 7 The real numbers inherit complex addition. Indeed, the inclusion map i : \mathbb{R} \to \mathbb{C} by x \mapsto x + 0i allows you to view any real number as a complex number in a canonical way. That is, when you add a real to a complex, you are effectively using complex addition. So, to answer the question directly: if you are being really ... 6 2^{\sin x } \cos x +1 = 1 \implies 2^{\sin x } \cos x = 0 \implies 2^{\sin x}=0 \text{ or } \cos x =0. Can you deduce the solution from here? 6 On the unit circle, \theta is the length of the arc (as well as the angle extended by that arc). (Thus, perimeter of the unit circle is 2\pi). Whereas, \cos\theta is the length of the X intercept, and \sin\theta is the length of the Y intercept. Look at the following diagram: You can now easily visualize that when Point P approaches closer to ... 6 HINT: Notice, we know that\cos(\pi-\theta)=-\cos \theta\tag 1$$Let,$$\cos \theta=x\iff \theta =\cos^{-1}(x)$$Where, \color{blue}{-1\leq x\leq 1} Then substituting the value of \theta in (1), we get$$\cos (\pi-\cos^{-1}(x))=-\cos(\cos^{-1}(x))\cos (\pi-\cos^{-1}(x))=-x\pi-\cos^{-1}(x)=\cos^{-1}(-x)$$... 6 Hint: The solutions of$$\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)are also solutions of \begin{align*} 7-4\sqrt{2}\sin x&=4\cos^2 x-4\sqrt{2}\cos x\tan x+2\tan^2 x\\ 7-4\sqrt{2}\sin x&=4\cos^2 x-4\sqrt{2}\sin x+2\tan^2 x \end{align*} Last equation is equivalent to4\cos^2 x+2\tan^2 x -7=0...(1)$$Let t=\cos^2 x, so (1) can be seen as ... 6$$ \vartheta=\arctan\frac{1}{2},\qquad \cos\vartheta = \frac{2}{\sqrt{5}}.$$6 The integral$$I = \int_{0}^{\pi/2} \sqrt{\sin(2\theta)} \, \sin\theta \, d\thetais evaluated by making use of the Beta function. This is seen as follows. \begin{align} I &= \int_{0}^{\pi/2} \sqrt{\sin(2\theta)} \, \sin\theta \, d\theta \\ &= \sqrt{2} \, \int_{0}^{\pi/2} \sin^{3/2}(\theta) \, \cos^{1/2}(\theta) \, d\theta \\ &= \sqrt{2} \cdot ... 5 If you are interested in knowing how to do this without using the Beta function, try the following steps. But I'm not going to write it out in full because it would take too long. Call the integral I. First do integration by parts, and we find thatI=\int_0^{\frac {\pi}{2}}\frac{\cos 2\theta\cos \theta}{\sqrt{\sin 2\theta}}d\theta$$Now add this form ... 5 The complex exponential function needs some definition. There are several ways one could define it: Power series As a solution to a differential equation As the unique holomorphic extension of the real exponential function to the whole plane Via the limit definition \lim_{n \to \infty} (1+\frac{z}{n})^n Via Euler's formula ... One can take any one of ... 4 It is enough to consider:$$ f(x)= \sin^3(x)-x^3\cos^2(x) $$and prove it is an increasing function over I=\left(0,\frac{\pi}{2}\right) by studying:$$ g(x)=\frac{f'(x)}{\cos(x)} = -3x^2\cos(x)+2x^3\sin(x)+3\sin^2(x).$$g(x)> 0 follows from \frac{\sin x}{x}> 1-\frac{x^2}{6} and \cos(x)< 1 over I. 4$$2\sin^2x=2\cos^2x-\sqrt3\iff2\sin^2x=2(1-\sin^2x)-\sqrt3\iff\sin^2x=\dfrac{2-\sqrt3}4=\dfrac{(\sqrt3-1)^2}{8}$$As \sin x<0,\sin x=-\dfrac{\sqrt3-1}{2\sqrt2}=\dfrac12\cdot\dfrac1{\sqrt2}-\dfrac{\sqrt3}2\cdot\dfrac1{\sqrt2}=\sin\left(\dfrac\pi6-\dfrac\pi4\right) x=n\pi+(-1)^n\left(\dfrac\pi6-\dfrac\pi4\right) where n is any integer 4 I think its conceptually cleaner to have both \cos and \sin as distinct notations, for the following reason: in my opinion, its a bit of a "coincidence" that each of the functions \{\cos,\sin\} can be described as translations of the other. My preferred definition of these two functions is the following: first, position yourself at (1,0) on the unit ... 4 A slightly different approach: With the change of variables u=\cos\theta, you arrive at$$\int_0^{\pi/2}\sqrt{\sin2\theta}\sin\theta\,d\theta=\sqrt2\int_0^1 u^{1/2}(1-u^2)^{1/4}\,du$$Another substitution, \sqrt t=u, yields$$\begin{align*}\sqrt2\int_0^1 t^{1/4}(1-t)^{1/4}\left(\frac{1}{2}t^{-1/2}\right)\,dt&=\frac{\sqrt2}{2}\int_0^1 ...

4

We know that $\tan (θ)=\frac{y}{x}$. Thus, for $\tan^{-1}(1/2)$, we have a right triangle whose $x$ and $y$ values are $2$ and $1$, respectively. Pythagorean theorem in terms of $x$, $y$, and $r$ is written as $x^2+y^2=r^2$. Simply plug in the values $x=1$ and $y=2$ from earlier to get the following: $$1^2+2^2=r^2$$ $$r^2=5$$ $$r=\sqrt5$$ Now, we ...

4

The identity you quote as being the textbook's answer - $|\cosh(x+i y)|^2 = \cos(x)^2 + \sinh(y)^2$ - is false in the case $y=0$ for all $x \not = 0$. Your identity and proof are correct. It is the case that $|\cosh(x+i y)|^2 = \sinh(x)^2 + \cos(y)^2$.

3

Hint: $$\cos5x+i\sin5x=\left(\cos x+i\sin x\right)^{5}=$$$$\cos^{5}x-10\cos^{3}x\sin^{2}x+5\cos x\sin^{4}x+i\left(5\cos^{4}x\sin x-10\cos^{2}x\sin^{3}x+\sin^{5}x\right)$$ (De Moivre)

3

Yes, that equation is completely wrong! $\tan(x)= \frac{\sin(x)}{\cos(x)}$. It is $\cot(x)$ that is $\frac{\cos(x)}{\sin(x)}$.

3

(Every real number is a complex number, e.g. $3=3+0i$, so adding a complex number and a real number is just adding two complex numbers.) With sine and cosine you should think in this context of circles rather than triangles. A first step toward understanding this is to realize this: Multiplying by $i$ means rotating $90^\circ$ counterclockwise. Doing ...

3

After using C-S we need to prove that \begin{align*} &\displaystyle\frac{9}{6-2\cos\gamma\cos\left(\alpha-\beta\right)+1-2\cos^2\gamma}\geq\frac{6}{5} \\ \iff&\displaystyle4\cos^2\gamma+4\cos\gamma\cos\left(\alpha-\beta\right)+1\geq0 \\ ...

3

You could do something like this, $$\frac{\sin\theta}{\cos\theta}\sin^2\theta = M$$ where M is your RHS , i am assuming constant. Let , $\alpha = \sin\theta$ then the equation becomes, $$\frac{\alpha^3}{\sqrt{1-\alpha^2}} = M$$ squaring both sides we have, $$\alpha^6= M^2(1-\alpha^2)$$ $$\alpha^6 + M^2\alpha^2 - M^2= 0$$ then you can solve for ...

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