Why do we need angle values greater than 90 degrees?

[Frigus]

I can't understand how can we assign values greater than 90 to trigonometric functions as right angle triangle can't exist if one angle is more than 90 degree. For example if I say sin 30 according to me it means that ratio of perpendicular and hypotenuse is 1/2 at 30 degree but how can we say something like this in angles greater than 120.

 

[fresh_42]

This is because we do not consider only (right) triangles, but the full circle instead which we divide into degrees. E.g. look at a compass and how pilots and captains measure their direction. And even in triangles, there are triangles with angles greater than 90° or 120°, and we also consider the outer angles, the complementary angles to the inner ones.

 

[PeroK]

I can't understand how can we assign values greater than 90 to trigonometric functions as right angle triangle can't exist if one angle is more than 90 degree. For example if I say sin 30 according to me it means that ratio of perpendicular and hypotenuse is 1/2 at 30 degree but how can we say something like this in angles greater than 120.

It's called a generalisation. Imagine the unit circle and start with your right-angle triangle in the first quadrant. You notice that:

$x = \cos \theta \$ and $y = \sin \theta$

As you continue round the circle, you could extend your definition of sine and cosine by taking these equations to define $\sin \theta$ and $\cos \theta$.

And then you have something even more useful than restricting yourself to angles less than $\pi/2$.

 

[WWGD]

You can even define a full coordinate system, polar coordinates, using sin, cos.

 

[HallsofIvy]

IF you are only using the "trig functions" on right triangles then there is no reason to use angles greater than 90 degrees. But generalizations of the trig functions (sometimes renamed "circular functions") are very useful as "periodic functions" modeling repetitive phenomena. As functions, we want them defined for all real numbers.

 

[Stephen Tashi]

I can't understand how can we assign values greater than 90 to trigonometric functions as right angle triangle can't exist if one angle is more than 90 degree.

Have you studied trigonometry as it is defined using the unit circle? If so, you understand how it is done. Perhaps your question is why it is done. Do you want to know why defining the trigonometric functions for all angles is useful?

 

[Agent Smith]

Have you studied trigonometry as it is defined using the unit circle? If so, you understand how it is done. Perhaps your question is why it is done. Do you want to know why defining the trigonometric functions for all angles is useful?

Yes please.

Perhaps @Frigus noticed that $\tan \frac{\pi}{2} = \tan 90^{\text{o}}$ still remains undefined, even when generalizing trig functions with a unit circle. The same problem we have with a right triangle with two $90^{\text{o}}$ angles we have with unit-circle-based definition of $\text{tangent}$. A "right triangle" with two $90^{o}$ cannot exist. How can there be "right triangles" with obtuse angles?

 

[Mark44]

Yes please.

Perhaps @Frigus noticed that $\tan \frac{\pi}{2} = \tan 90^{\text{o}}$ still remains undefined, even when generalizing trig functions with a unit circle. The same problem we have with a right triangle with two $90^{\text{o}}$ angles we have with unit-circle-based definition of $\text{tangent}$. A "right triangle" with two $90^{o}$ cannot exist. How can there be "right triangles" with obtuse angles?

The aim of extending or generalizing right triangle trig is not to provide definitions for such expressions as $\tan(\pi/2)$ or $\csc(0)$. It is to be able to define values for the six trig functions for all real angle values, not just those between 0 and 90°. Of course, the trig functions that are defined in terms of division (tangent, cotangent, secant, cosecant) have domains that don't permit certain values.

Regarding your comment about the impossibility of a right triangle with two right angles, that's true if we're talking only about plane surfaces. However, without this limitation it's possible to have a right triangle with three right angles. Suppose you're standing at the north pole. You walk due south for one mile, and then turn left, making a 90° angle. Head due east for one mile and turn left again, making another 90° angle. Head due north for one mile to reach your starting point at the north pole. Your path determines an equilateral triangle all of whose angles are 90°.

 

[Agent Smith]

@Mark44 , muchas gracias.

$\sin (\theta) = \sin (180^o - \theta)$

$\cos (\theta) = \cos (360^o - \theta)$

$\tan (\theta) = \tan (180^o + \theta)$

Would I be correct to say that the trig function values for obtuse and reflex angles are equal to the trig function values of their corresponding acute angles.

 

[Mark44]

Would I be correct to say that the trig function values for obtuse and reflex angles are equal to the trig function values of their corresponding acute angles.

This is a bit too general. You can answer your own question by using the trig identities for sums and differences of angles. From them you should be able to see that the sine of an angle and its supplement are equal, but the cosine of an angle and its supplement differ in sign. IOW $\cos(\theta) = -\cos(\pi - \theta)$.

 

[Agent Smith]

You can answer your own question

I wish!

I was only trying to give the OP an idea of what I felt was some kind of pair-matching between reflex and obtuse angles and their corresponding acute angles (supplementary i.e. sum to 180 degrees and sum-to-360 degrees) in re their trig function values.

 

[Mark44]

You can answer your own question by using the trig identities for sums and differences of angles.

I wish!

Why not? The sum and difference trig identities are in every textbook on trig or can easily be found online, like on wikipedia.

I was only trying to give the OP an idea of what I felt was some kind of pair-matching between reflex and obtuse angles and their corresponding acute angles (supplementary i.e. sum to 180 degrees and sum-to-360 degrees) in re their trig function values.

I'm not sure the OP is still paying attention in a thread that is almost five years old and who hasn't been heard of for more than a year.

 

[Agent Smith]

@Mark44 I find it intriguing that generalizing trig functions with a unit circle gave meaning to trig functions for all angles (not just acute angles) except $\tan \frac{\pi}{2}$.
$\cos 0 = 1$ (a degenerate triangle T1)
$\sin \frac{\pi}{2} = 1$ (another degenerate triangle T2)
$\tan \frac{\pi}{2} = \text{undefined}$ (the same degenerate triangle T2)