trigonometric functions

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k a May Highlights and 2025 AoPS Online Class Information

jlacosta 0

Thursday at 11:16 PM

May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies

jlacosta

Thursday at 11:16 PM

0 replies

Algebra problem

Deomad123 0

9 minutes ago

Let $n$ be a positive integer.Prove that there is a polynomial $P$ with integer coefficients so that $a+b+c=0$ ,then $a^{2n+1}+b^{2n+1}+c^{2n+1}=abc[P(a,b)+P(b,c)+P(a,c)]$ .

0 replies

Deomad123

9 minutes ago

0 replies

9 Rate my profile

Evanlovemath 2

N 23 minutes ago by Evanlovemath

https://artofproblemsolving.com/alcumus/profile/613246 Not doing Alcumus right now, to many classes :(

2 replies

Evanlovemath

25 minutes ago

Evanlovemath

23 minutes ago

trigonometric functions

VivaanKam 11

N 2 hours ago by aok

Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!

11 replies

VivaanKam

Apr 29, 2025

aok

2 hours ago

1201 divides sum of powers

V0305 1

N 2 hours ago by vincentwant

(Source: me) Prove that for all positive integers $n$ , $1201 \mid 2^{2^n} + 59^{2^n} + 61^{2^n}$ .

1 reply

V0305

3 hours ago

vincentwant

2 hours ago

Interesting geometry

polarLines 5

N 3 hours ago by Mathworld314

Let $ABC$ be an equilateral triangle of side length $2$ . Point $A'$ is chosen on side $BC$ such that the length of $A'B$ is $k<1$ . Likewise points $B'$ and $C'$ are chosen on sides $CA$ and $AB$ . with $CB'=AC'=k$ . Line segments are drawn from points $A',B',C'$ to their corresponding opposite vertices. The intersections of these line segments form a triangle, labeled $PQR$ . Prove that $\Delta PQR$ is an equilateral triangle with side length ${4(1-k) \over \sqrt{k^2-2k+4}}$ .

5 replies

1 viewing

polarLines

May 20, 2018

Mathworld314

3 hours ago

Showing that certain number is divisible by 13

BBNoDollar 3

N 3 hours ago by Shan3t

Show that 3^(n+2) + 9^(n+1) + 4^(2n+1) + 4^(4n+1) is divisible by 13 for every n natural number.

3 replies

BBNoDollar

5 hours ago

Shan3t

3 hours ago

Sum of digits is 18

Ecrin_eren 4

N 4 hours ago by maxamc

How many 5 digit numbers are there such that sum of its digits is 18

4 replies

Ecrin_eren

Today at 1:10 PM

maxamc

4 hours ago

Inequality

tom-nowy 0

5 hours ago

Let $0<a,b,c,<1$ . Show that
$\frac{3(a+b+c)}{a+b+c+3abc} > \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} .$

0 replies

tom-nowy

5 hours ago

0 replies

Logarithm of a product

axsolers_24 2

N 5 hours ago by axsolers_24

Let $x_1=97 ,$ $x_2=\frac{2}{x_1} ,$ $x_3=\frac{3}{x_2} ,$ $... ,$ $x_8=\frac{8}{x_7}$
then
$\log_{3\sqrt{2}} \left(\prod_{i=1}^8 x_i-60\right)$

2 replies

axsolers_24

Today at 10:42 AM

axsolers_24

5 hours ago

Inequalities

sqing 1

N 5 hours ago by sqing

Let $a,b>0 , a^2 + 2b^2 = a + 2b$ . Prove that $\sqrt{\frac{a}{b( a+2)}} + \sqrt{\frac{b}{a(2b+1)}} \geq \frac {2}{\sqrt{3}}$ Let $a,b>0 , a^3 + 2b^3 = a + 2b$ . Prove that $\sqrt[3]{\frac{a}{b( a+2)}} + \sqrt[3]{\frac{b}{a(2b+1)}} \geq \frac {2}{\sqrt[3]{3}}$

1 reply

sqing

5 hours ago

sqing

5 hours ago

Coprime sequence

Ecrin_eren 4

N 5 hours ago by Pal702004

"Let N be a natural number. Show that any two numbers from the following sequence are coprime:

2^1 + 1, 2^2 + 1, 2^4+ 1,2^8+1 ..., 2^(2^N )+ 1."

4 replies

Ecrin_eren

Thursday at 8:53 PM

Pal702004

5 hours ago

Hard Inequality

William_Mai 0

5 hours ago

Given $a, b, c \in \mathbb{R}$ such that $a^2 + b^2 + c^2 = 1$ .
Find the minimum value of $P = ab + 2bc + 3ca$ .

Source: Pham Le Van

0 replies

William_Mai

5 hours ago

0 replies

Find the minimum

Ecrin_eren 5

N 6 hours ago by Jackson0423

The polynomial is given by P(x) = x^4 + ax^3 + bx^2 + cx + d, and its roots are x1, x2, x3, x4. Additionally, it is stated that d ≥ 5.Find the minimum value of the product:

(x1^2 + 1)(x2^2 + 1)(x3^2 + 1)(x4^2 + 1).

5 replies

Ecrin_eren

Thursday at 9:03 PM

Jackson0423

6 hours ago

Inequalities

sqing 8

N 6 hours ago by sqing

Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$(1-abc) (1-a)(1-b)(1-c) \ge 208$ $(1+abc) (1-a)(1-b)(1-c) \le -224$ $(1+a^2b^2c^2) (1-a)(1-b)(1-c) \le -5840$

8 replies

sqing

Jul 12, 2024

sqing

6 hours ago

trigonometric functions

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156 posts

#1 h Apr 29, 2025, 8:29 PM

Y by

Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!

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222 posts

Lijin

#2 h Apr 29, 2025, 8:43 PM

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Are you talking about graphing them or just the basic ratios?

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1265 posts

Yiyj1

#4 h Apr 29, 2025, 8:57 PM

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Basic ratios: draw a right triangle. I'm terrible at asy so i can't draw one D:. Anyways, label one of the angles $\theta$ . Then, label the hypotenuse with $H$ , the leg adjacent to $\theta$ as $A$ (for adjacent) and the other leg $O$ (for opposite). Then, just remember this: SOH CAH TOA: $\sin\theta=\dfrac{O}{H}, \cos\theta=\dfrac{A}{H}, \tan\theta=\dfrac{O}{A}.$ Then there are like $\sec, \csc, \cot$ , which are the reciprocals of $\cos, \sin, \tan$ . IMPORTANT: $\sec$ is the reciprocal of $\cos$ and $\csc$ is the reciprocal of $\sin$ , not the other way around.

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332 posts

aok

#5 h Apr 29, 2025, 10:08 PM

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To solve things such as $\sin 150$ or $\cos 270$ you can draw a circle of radius 1 around the origin, rotate a line from the positive x line counterclockwise by $\theta^\circ$ to form a new line. To solve for cos,sin, and tan $\theta,$ where the $(x,y)$ hits circle after rotation and $x = \cos\theta$ , $y = \sin\theta$ , and $\frac{y}{x} = \tan\theta.$

This post has been edited 4 times. Last edited by aok, Apr 29, 2025, 10:48 PM

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156 posts

#6 h Apr 30, 2025, 6:25 PM

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Yiyj1 wrote:

Basic ratios: draw a right triangle. I'm terrible at asy so i can't draw one D:. Anyways, label one of the angles $\theta$ . Then, label the hypotenuse with $H$ , the leg adjacent to $\theta$ as $A$ (for adjacent) and the other leg $O$ (for opposite). Then, just remember this: SOH CAH TOA: $\sin\theta=\dfrac{O}{H}, \cos\theta=\dfrac{A}{H}, \tan\theta=\dfrac{O}{A}.$ Then there are like $\sec, \csc, \cot$ , which are the reciprocals of $\cos, \sin, \tan$ . IMPORTANT: $\sec$ is the reciprocal of $\cos$ and $\csc$ is the reciprocal of $\sin$ , not the other way around.

So like this?

$[asy] draw((0,0)--(3,0)--(0,2)--cycle); label("$\theta$", (2.7,0.1),W); label("$A$", (1.5,0), S); label("$O$", (0,1.205), W); label("$H$", (1.2,1.1), NE); [/asy]$

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156 posts

#7 h Apr 30, 2025, 6:26 PM

Y by

That’s cool! So if you have the lengths of a triangle you can find its angles?

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156 posts

#8 h Apr 30, 2025, 6:31 PM

Y by

aok wrote:

To solve things such as $\sin 150$ or $\cos 270$ you can draw a circle of radius 1 around the origin, rotate a line from the positive x line counterclockwise by $\theta^\circ$ to form a new line. To solve for cos,sin, and tan $\theta,$ where the $(x,y)$ hits circle after rotation and $x = \cos\theta$ , $y = \sin\theta$ , and $\frac{y}{x} = \tan\theta.$

are they like polar quardinits ?

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156 posts

#9 h Apr 30, 2025, 6:34 PM

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but the wouldn't $\cos x$ have 2 values because on a circle there are two quordinates with the same $x$ position?

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2988 posts

#10 h Apr 30, 2025, 6:38 PM

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Yes, you can find the side lengths of a triangle given its angles. (If it is right, you can just use the standard ratio definitions of $\sin, \cos, \tan$ and use inverse trigonometric functions, and if it is not right, then you can use the Law of Cosines to find each angle.)

The point in rectangular coordinates $(\cos(\theta), \sin(\theta))$ corresponds to the point in polar coordinates $(1, \theta),$ i.e. $(\cos(\theta), \sin(\theta))$ is the point on the unit circle at an angle of $\theta$ radians counterclockwise of the positive $x$ -axis.

Yes, the equation $\cos(x)=a$ has two solutions in $[0, 2\pi)$ for all $-1<a<1.$

Also, reminder that you can combine all of your questions into the same post.

This post has been edited 1 time. Last edited by lpieleanu, Apr 30, 2025, 6:39 PM

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332 posts

aok

#11 h Apr 30, 2025, 7:12 PM

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that is correct, cos x = a has 2 solutions (generally)

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aok

#12 h Yesterday at 12:43 AM

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for x btw

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aok

#13 h 2 hours ago

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VivaanKam wrote:

That’s cool! So if you have the lengths of a triangle you can find its angles?

Correct, use the opposite of those functions.

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