Tag Info

Hot answers tagged

16

To quote a previous answer of mine: Robison, "A new approach to circular functions, π, and lim sin(x)/x", Math. Mag. 41.2 (March 1968), 66–70 [jstor]. In this paper it is shown that the addition law for cosine (and a couple other simple assumptions) uniquely determines cosine and sine. (This is the paper I cite most often on StackExchange.) If ...


11

From what you do not want that pretty much leaves functional equations. There seems to be a system of two functional equations for sine and cosine: (link) $$ \Theta(x+y)=\Theta(x)\Theta(y)-\Omega(x)\Omega(y) \\ \Omega(x+y)=\Theta(x)\Omega(y)+\Omega(x)\Theta(y) $$


11

Roughly, they are the same definition of tangent: $$\begin{align} \require{cancel} \tan = \frac{\sin}{\cos} &= \frac{\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)} {\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)} \\ &= \left(\frac{\text{opposite}}{\text{hypotenuse}}\right) \div ...


10

Note: in a triangle ABC, we have $A+B+C=\pi=180^o$ Given $$\sin B\sin C=\cos^2\frac{A}{2} \implies 2\sin B\sin C=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)-\cos(B+C)=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)-\cos(180^o-A)=2\cos^2\frac{A}{2} $$ $$\implies \cos(B-C)+\cos A=2\cos^2\frac{A}{2} $$$$\implies \cos(B-C)+2\cos^2\frac{A}{2}-1=2\cos^2\frac{A}{2}$$ ...


7

Lets consider the indefinite integral, $$ \int \sec(x)\tan(x) dx = \int \frac{\sin(x)}{\cos^2(x)} dx$$ We can then perform a $u$ substitution with $u=\cos(x)$ and $du = -\sin(x) dx $ obtaining, $$ \int \sec(x)\tan(x) dx = -\int \frac{1}{u^2} du= \frac{1}{u} + C = \frac{1}{\cos(x)} +C = \sec(x) + C$$


7

drop a perpendicular from $A$ to $DC$ and call the foot of the perpendicular $E.$ then $$DE = EC = \cos \alpha, DC = 2 \cos \alpha.$$ now look at the right triangle $CBD$ with hypotenuse $2\cos \alpha$ and the $\angle BDC = \alpha,$ therefore the opposite side $BC$ is $$ BC = 2\cos \alpha \sin \alpha.$$ you can also look at the right angle triangle ...


7

Let $$f(\theta)=\cos ^2\theta+\cos^2 (\theta+1^{\circ})+\cos^2(\theta+2^{\circ})+...... +\cos^2(\theta+179^{\circ})$$ Then $$f(\theta+1)-f(\theta)=\cos^2(\theta+180^{\circ})-\cos ^2\theta =0$$ This shows that $f(\theta)$ is periodic with period $1$. In particular $$f(\theta)=f(\theta+90)$$ Show now that $$f(\theta+90)=\sin ^2\theta+\sin^2 ...


7

$$\cos(2x)+1+3\sin x=0$$ now let's substitute $\cos (2x)=1-2\sin^2x$ $2-2\sin^2x+3\sin x=0$ And then, if we denote $y=\sin x$ we get a quadratic equation $$2-2y^2+3y=0$$


6

Note that $-1\le \sin \theta \le 1\implies 11\le 12-\sin \theta \le 13$ Also $-1\le\cos \theta \le 1\implies -1\le \cos 2\theta \le 1$ Can you take it from there?


6

Hint: Here's something for you to think about. By the Chain Rule $$\frac{d}{dx} \left(\cos (x)\right)^{-1} = (-1) (\cos (x))^{-2} \dot \, (-\sin (x)) = \frac{1}{\cos x} \dot\, \frac{\sin x}{\cos x}$$ Edit: In order to find $\int \sec x \tan x dx $. Write it as $$\frac{\sin x}{\cos^2 x}$$ and make the substitution $u = \cos x$.


5

The other solutions come from solving $\sin\theta=0$. You can't just cancel the $\sin\theta$ just because it's a common factor, without considering the possibility it could be zero. It's like when you solve the quadratic equation $x^2-3x=0$. Would you say that the only solution is $x=3$?


5

In this special case, you can use sum of sine: $$\sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$$ So if your $20,40$ are in degrees, $$\sin 20^\circ +\sin 40^\circ=2\sin 30^\circ\cos 10^\circ=\cos 10^\circ$$ If $\cos 10^\circ=\sin x$, then $x=80^\circ$ or $100^\circ$, or plus integer multiples of $360^\circ$.


5

$$\begin{align} \sum_{k=0}^{179}\cos^2(\theta+k\cdot 1^\circ) &= \sum_{k=0}^{89}\cos^2(\theta+k\cdot 1^\circ) + \cos^2(\theta+k\cdot 1^\circ + 90^\circ) \\ &= \sum_{k=0}^{89}\cos^2(\theta+k\cdot 1^\circ) + \sin^2(\theta+k\cdot 1^\circ)\\ &= \sum_{k=0}^{89} \; 1 \\ &= 90 \end{align}$$


5

If $x=0$ and $\sin y$ is transcendental - say $\sin y = \pi/4$ and $\cos y>0$ - then $\sin(x+z)=\sin z$ being rational means $\cos z$ is algebraic, so $$\sin(y+z)=\sin y\cos z + \sqrt{1-\sin^2 y} \sin z$$ being rational means $\sin y$ is algebraic. So there is no such $z$ in this case. Indeed, for any $x$ there are at only countably many $y$ so that ...


4

Hint: $\color{#0ae}{(1-\sin^2 x)}-\color{#0b4}{(\cos 2x)}=\color{#0ae}{(\cos^2 x)} -\color{#0b4}{(\cos^2x-\sin^2x)}=\sin^2x =\frac{1}{2}$ $\iff \sin x=\pm\frac{\sqrt{2}}{2}$. You should be able to solve this.


4

$\cos 4x=\cos 3x$ $\implies 4x=2n\pi\pm 3x$ where $n\in \mathbb{Z}$ Taknig positive sign we have $4x=2n\pi + 3x $ $\implies x=2n\pi ~ ;n\in \mathbb{Z}$ Taking negative sign we have $4x=2n\pi - 3x$ $\implies 7x=2n\pi$ $\implies x= \frac{2n\pi}{7} ~; n\in \mathbb{Z}$


4

Yes, there are nice geometric explanations of the derivative formulas for all six basic trig functions, which ought to be much more widely known. (I hesitate to use the word "proof" for an argument that uses infinitesimals.) They are all based on the following fact about isosceles triangles: For an isosceles triangle with small vertex angle $d\theta$, the ...


4

Assume as a contraddiction that there exists $(a,b)\neq(0,0)$ such that $\forall x\in[-\pi,\pi]\ a\sin x+b\cos x=0$ Then, for all $x\in[-\pi,\pi]$ it would hold $$\begin{cases}\cos^2x+\sin^2x=1\\a\sin x+b\cos x=0\end{cases}$$ But the system $$\begin{cases}t^2+s^2=1\\at +bs=0\end{cases}$$ has exactly $2$ solutions, which would imply that $\cos x$ and $\sin ...


4

Try using an identity. $\sin 2\alpha = \sin \alpha + \alpha = \sin \alpha \cos \alpha + \sin\alpha\cos\alpha = 2\sin\alpha\cos\alpha $


3

The first piece of the definition says it all. $\arctan$ is a map $\mathbb R\to(-\frac\pi 2,\frac \pi 2)$, so the other adjustments are an attempt to return the correct argument for the point $(x,y)$; this argument can lie in $(-\pi,\pi]$. For example, for the complex number $-1-i$ (taking $x=y=-1$), the argument should be $-\frac {3\pi}{4}$. But ...


3

With an obvious shorthand notation, we can solve the equation for $\cos C$: $$c^2+2abc+a^2+b^2-1=0,$$ $$c=-ab\pm\sqrt{a^2b^2-a^2-b^2+1}=-ab\pm\sqrt{(1-a^2)(1-b^2)}.$$ Then $$\cos C=-\cos A\cos B\pm \sin A\sin B=-\cos(A\pm B)$$ and $$\pm C=\pi\pm A\pm B.$$


3

You can think of $\arctan(2)$ as an angle whose tangent is $2$. To visualize such an angle, draw a right triangle and mark an acute angle. Then opposite the marked angle have a side (leg) of length $2$, and adjacent to the marked angle have a side (leg) of length $1$. The marked angle now represents $\arctan(2)$. You can then fill in the hypotenuse ...


3

The equilateral triangle is the only such case. Let $A$ and $B$ be two angles, and $a$ and $b$ be the respective opposite sides. Then $$ \frac{\sin A}{\sin B} = \frac a b $$ so if your ratios match, then $$ \frac{\sin A}{\sin B} = \frac a b $$ $$ \frac{\sin A}{A} = \frac{\sin B}{B} $$ So to have a triangle where $A \neq B$ you require two distinct ...


3

The expression $\sec x\tan x$ can be written $$ \frac{1}{\cos x}\frac{\sin x}{\cos x}=\frac{\sin x}{\cos^2 x} =-\frac{-\sin x}{\cos^2 x}=-\frac{f'(x)}{f(x)^2} $$ where $f(x)=\cos x$. Consider, for a generic differentiable function $f$, $$ g(x)=\frac{1}{f(x)}. $$ By the chain rule $$ g'(x)=-\frac{f'(x)}{f(x)^2}. $$ In the special case of $f(x)=\cos x$, we see ...


3

HINT: Using Basic trigonometry intuition, we have $4x=2m\pi\pm3x$ Considering the '+' sign, $x=2m\pi$ Considering the '-' sign, $x=2m\pi/7$ where $0\le m\le6$ But $\cos(2\pi-y)=\cos y$ So, $\cos\dfrac{(7-r)2\pi}7=\cos\left[2\pi-\dfrac{2r\pi}7\right]=\cos\dfrac{2r\pi}7$ Set $r=1,2,3$


3

you are almost there. $$2\sin^2 \theta + \sin \theta - 12=0 $$ so $$\sin \theta = \frac{-1 \pm \sqrt{49}}{4} = -2, \frac32$$ neither of them leads to a solution for $\theta.$ we should have seen this from the equation itself. reason is $$1 \ge |\cos 2\theta| = |12 - \sin \theta | \ge 11$$ a contradiction.


3

Hint: So far so good: $$2\sin^2\theta + \sin\theta - 12= 0 \implies \sin \theta = \frac{-1\pm\sqrt{1^2-4\cdot 2 \cdot(-12)}}{2\cdot 2}.$$ Find two possible values for $\sin \theta$. Then find the possible values for $\theta$ (will there be any?).


3

Let $\cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})=y$ $\iff \sin \theta +\sqrt{\sin ^2 \theta +3}=y\sec\theta$ $\iff \sqrt{\sin ^2 \theta +3}=y\sec\theta-\sin \theta$ Squaring we get $\sin ^2 \theta +3=y^2(1+\tan^2\theta)+\sin^2\theta-2y\tan\theta$ $\iff y^2(\tan^2\theta)-2y(\tan\theta)+y^2-3=0$ As $\tan\theta$ is real, the discriminant must be ...


3

For simplicity's sake, write $u = \frac{\pi}2 - 4x$; then the problem you are trying to solve becomes $$\cos y = \cos u$$ Now let's think about when it's possible for $y$ and $u$ to have the same cosine. If you interpret $y$ as an angle (measured in radians) that locates a point on a unit circle, then $\cos y$ tells you the $x$-coordinate of that point. ...


3

The conclusion from the back-and-forth discussion in the comments appears to be that with regard to your concepts of "already squared" and "should be squared in the future" the superscript $2$ in expressions such as $$\sin^2 \theta, \quad 5^2,\quad \mbox{or}\quad h^2$$ means "should be squared in the future," with no exceptions. As explained in comments and ...



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