trigonometric functions

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VivaanKam

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VivaanKam

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

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

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Lijin

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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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Yiyj1

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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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aok

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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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VivaanKam

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VivaanKam

#6 h Wednesday at 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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VivaanKam

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VivaanKam

#7 h Wednesday at 6:26 PM

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

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VivaanKam

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VivaanKam

#8 h Wednesday at 6:31 PM

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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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VivaanKam

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VivaanKam

#9 h Wednesday at 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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lpieleanu

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lpieleanu

#10 h Wednesday at 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, Wednesday at 6:39 PM

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aok

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aok

#11 h Wednesday at 7:12 PM

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

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aok

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aok

#12 h Today at 12:43 AM

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for x btw

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