By meaning we recognize that:\$\$ extcos alpha = frac extadjacent exthypotenuse.\$\$

If we want to apply the meaning to the situation in the photo below:

we have actually that:\$\$ extcos 90° = frac?h .\$\$How can I say the it is same to \$0\$ if ns don"t recognize anything around the various other two sides, or around the other two angles?

I have actually been maybe to constantly find a value, also without the unit circle, in instances like \$ extcsc 90°, extsec 0°\$, etc..., yet not in the above situation. Why?

Please, can you suggest me anything?

So, i make an addition also based upon suggestions provided.My main error was to begin to take into consideration the best angle, instead I have to start considering \$ heta = alpha°\$, and increse it till \$ heta = 90°\$, one side end up being smaller till zero, and the various other side end up being bigger till equal to \$h\$, therefore \$ extcos alpha = frac0h = 0\$

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

JB-FrancoJB-Franco
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Others have mentioned the such a best 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 made decision \$45^circ\$ since you can work this precisely with one isosceles ideal angled triangle of side length \$1,1,sqrt2\$

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

KarlKarl
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While the trigonometric features are initially characterized for the angle of a triangle (in radians) lock are extended to all genuine numbers, and eventually, to facility numbers. While their properties, such together the enhancement laws, room preserved, they eventually lose all connection with triangles.

In the case you give, that is clear the the adjacent side it s okay closer and also closer come \$0\$ as the angle it s okay closer come \$pi/2\$, so the cosine it s okay close come \$0,\$ yet you clearly can"t really have actually a triangle v two appropriate angles.

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answered Apr 22 "18 at 11:58

saulspatzsaulspatz
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There is a cons in her reasoning. When doing trigonometry (in a timeless way) we think about a best angled triangle. The value of \$cos(alpha)\$ is frequently geven as the proportion \$frac extadjacent side exthypotenuse.\$ This definition, however, is incomplete and also would not be found like this in netherlands school books for instance. Netherlands shool books would say something like:\$\$cos (alpha)=frac extaanliggende \$colorred extrechthoeks\$zijde exthypotenusa.\$\$ The important difference here is that "rechthoekszijde" can"t simply be any side that a triangle. A "rechte hoek" is a appropriate angle, and also a "rechthoekszijde" is a side of a best angled triangle that is surrounding to its ideal angle (so no the hypotenuse by definition).

This method that if you desire to deduce what \$cos(90^circ)\$ need to be, you need to make sure that the adjacent side is not just adjacent to the angle \$alpha\$, but likewise to the appropriate angle that the triangle.

N.B. Save in mind the this means that \$cos(90^circ)\$ deserve to only be "observed" in a triangle through two ideal angles. You deserve to interpret this in multiple ways: a degenerate triangle with one vertex in ~ the suggest at infinity; the border \$lim_h oinftyfrac exta exth\$; maybe some other way.

In any case, the an essential point right here is this:

When doing classical trigonometry, in common case yet definitely additionally in "extreme" cases, you require to specify adjacent side and also opposite side in such a way that one of two people of them deserve to never it is in the hypotenuse.

Below is a snapshot in i m sorry I try do demonstrate how you can use timeless trigonomtry to define the trigonometric features consistently outside \$<0,fracpi2)\$ = \$<0^circ,90^circ)\$.

For instance, if the angle gets bigger 보다 \$90^circ\$ then \$C\$ will come to lie below \$B\$ in stead of above and we consider "opposite" and also "hypotenuse" to be negative.