By meaning we recognize that:$$\\textcos \\alpha = \\frac\\textadjacent\\texthypotenuse.$$

If we want to use the definition to the situation in the picture below: we have that:$$\\textcos 90° = \\frac?h .$$How have the right to I say the it is same to $0$ if ns don\"t recognize anything around the other two sides, or about the other two angles?

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

Please, have the right to you suggest me anything?

So, ns make an enhancement also based upon suggestions provided.My main error to be to start to think about the ideal angle, instead I need to start considering $\\theta = \\alpha°$, and also increse it till $\\theta = 90°$, one side end up being smaller it spins zero, and the various other side end up being bigger till equal to $h$, as such $\\textcos \\alpha = \\frac0h = 0$ You are watching: What is the cosine of 90

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edited Apr 22 \"18 in ~ 18:40
JB-Franco
asked Apr 22 \"18 at 11:42 JB-FrancoJB-Franco
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While the trigonometric functions are initially identified for the angles of a triangle (in radians) they are extended to all real numbers, and eventually, to complicated numbers. While their properties, such as the addition laws, space preserved, they ultimately lose all link with triangles.

In the instance you give, that is clear the the nearby side gets closer and closer to $0$ as the angle gets closer come $\\pi/2$, so the cosine gets close come $0,$ yet you clear can\"t really have a triangle with two appropriate angles.

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answer Apr 22 \"18 at 11:58 saulspatzsaulspatz
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Others have actually mentioned that such a appropriate angle triangle does no exist.

As a compromise exactly how about

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

The addition formula is easily justified geometrically.

I determined $45^\\circ$ due to the fact that you deserve to work this exactly with one isosceles ideal angled triangle that side length $1,1,\\sqrt2$

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edited Apr 22 \"18 at 12:22
reply Apr 22 \"18 in ~ 11:49 KarlKarl
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There is a flaw in your reasoning. As soon as doing trigonometry (in a timeless way) we take into consideration a right angled triangle. The worth of $\\cos(\\alpha)$ is commonly geven as the proportion $\\frac\\textadjacent side\\texthypotenuse.$ This definition, however, is incomplete and would no 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 that \"rechthoekszijde\" can\"t just be any kind of side of a triangle. A \"rechte hoek\" is a ideal angle, and a \"rechthoekszijde\" is a side of a ideal angled triangle that is adjacent to its best angle (so no the hypotenuse through definition).

This means that if you desire to deduce what $\\cos(90^\\circ)$ should be, you need to make sure that the adjacent side is no just nearby to the edge $\\alpha$, but also to the best 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 best angles. You deserve to interpret this in many ways: a degenerate triangle v one vertex at the suggest at infinity; the border $\\lim_h\\to\\infty\\frac\\texta\\texth$; probably some other way.

In any case, the vital point below is this:

When doing classic trigonometry, in normal case but definitely additionally in \"extreme\" cases, you require to specify adjacent side and opposite side in together a method that either of them can never it is in the hypotenuse.

Below is a picture in which I shot do show how you deserve to use classical trigonomtry to define the trigonometric functions consistently external $<0,\\frac\\pi2)$ = $<0^\\circ,90^\\circ)$. For instance, if the angle it s okay bigger than $90^\\circ$ then $C$ will pertained to lie listed below $B$ in stead of over and we take into consideration \"opposite\" and also \"hypotenuse\" to be negative.