By an interpretation we know that:$$\textcos \alpha = \frac\textadjacent\texthypotenuse.$$

If we want to apply the meaning to the instance in the picture below:

we have actually that:$$\textcos 90° = \frac?h .$$How deserve to I say that it is equal to $0$ if i don"t know anything about the other two sides, or about the other two angles?

I have actually been able to always find a value, also without the unit circle, in situations like $\textcsc 90°, \textsec 0°$, etc..., however not in the above situation. Why?

Please, can you suggest me anything?

So, i make an enhancement also based on suggestions provided.My key error to be to start to think about the ideal angle, rather I need to start considering $\theta = \alpha°$, and also increse that till $\theta = 90°$, one side become smaller it rotates zero, and the other side end up being bigger till same to $h$, as such $\textcos \alpha = \frac0h = 0$

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edited Apr 22 "18 in ~ 18:40
JB-Franco
inquiry Apr 22 "18 in ~ 11:42

JB-FrancoJB-Franco
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While the trigonometric features are initially characterized for the angle of a triangle (in radians) they are prolonged to all actual numbers, and eventually, to facility numbers. While your properties, such together the enhancement laws, are preserved, they at some point lose all connection with triangles.

In the situation you give, the is clear that the adjacent side gets closer and also closer come $0$ as the angle gets closer come $\pi/2$, so the cosine gets close come $0,$ but you clear can"t really have a triangle v two best angles.

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answer Apr 22 "18 in ~ 11:58

saulspatzsaulspatz
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Others have actually mentioned the such a ideal angle triangle does not exist.

As a compromise just how about

$$\cos(45+45)=\cos(45)\cos(45)-\sin(45)\sin(45)=0$$

The addition formula is easily justified geometrically.

I determined $45^\circ$ because you deserve to work this precisely with one isosceles right angled triangle that side size $1,1,\sqrt2$

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edited Apr 22 "18 at 12:22
answered Apr 22 "18 at 11:49

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See more: What Type Of Bond Is Cacl2 A Covalent Bond? Is Cacl2 (Calcium Chloride) Ionic Or Covalent

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There is a defect in her reasoning. Once doing trigonometry (in a classical way) we consider a appropriate angled triangle. The value of $\cos(\alpha)$ is generally geven together the ratio $\frac\textadjacent side\texthypotenuse.$ This definition, however, is incomplete and also would not be found like this in dutch school publications for instance. Netherlands shool publications would speak something like:$$\cos (\alpha)=\frac\textaanliggende \colorred\textrechthoekszijde\texthypotenusa.$$ The important distinction here is the "rechthoekszijde" can"t simply be any type of side the a triangle. A "rechte hoek" is a ideal angle, and a "rechthoekszijde" is a next of a best angled triangle that is nearby to its best angle (so no the hypotenuse by definition).

This means that if you desire to deduce what $\cos(90^\circ)$ must be, you have to make sure that the adjacent side is no just surrounding to the angle $\alpha$, but additionally to the ideal angle the the triangle.

N.B. Store in mind the this method that $\cos(90^\circ)$ have the right to only it is in "observed" in a triangle v two appropriate angles. You deserve to interpret this in many ways: a degenerate triangle through one vertex at the suggest at infinity; the limit $\lim_h\to\infty\frac\texta\texth$; probably some other way.

In any type of case, the key point here is this:

When doing timeless trigonometry, in normal case yet definitely additionally in "extreme" cases, you need to specify adjacent side and also opposite side in together a way that either of them have the right to never it is in the hypotenuse.

Below is a picture in i m sorry I shot do show how you have the right to use classic trigonomtry to specify the trigonometric features consistently external $<0,\frac\pi2)$ = $<0^\circ,90^\circ)$.

For instance, if the angle gets bigger than $90^\circ$ climate $C$ will involved lie listed below $B$ in stead of over and we think about "opposite" and also "hypotenuse" to it is in negative.