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Welcome to arccos -(sqrt6 + sqrt2)/4, our post aboutthe arccosine of -(sqrt6 + sqrt2)/4.
For the inverse trigonometric function of cosine -(sqrt6 + sqrt2)/4 we usually employ the abbreviation arccos and write it as arccos -(sqrt6 + sqrt2)/4 or arccos(-(sqrt6 + sqrt2)/4).
If you have been looking for what is arccos -(sqrt6 + sqrt2)/4, either in degrees or radians, or if you have been wondering about the inverse of cos -(sqrt6 + sqrt2)/4, then you are right here, too.
In this post you can find the angle arccosine of -(sqrt6 + sqrt2)/4, along with identities.
Read on to learn all about the arccos of -(sqrt6 + sqrt2)/4, and note that the term -sqrt(6)/4 – sqrt(2)/4 is approximately -0.96592582 as a decimal number.
Arccos of -(sqrt6 + sqrt2)/4
If you want to know what is arccos -(sqrt6 + sqrt2)/4 in terms of trigonometry, check out the explanations in the last paragraph; ahead in this section is the value of arccosine(-(sqrt6 + sqrt2)/4):
arccosine -(sqrt6 + sqrt2)/4 = 11pi/12 rad = 165 °
arccosine of -(sqrt6 + sqrt2)/4 = 11pi/12 radians = 165 degrees
The arccos of -(sqrt6 + sqrt2)/4 is 11pi/12 radians, and the value in degrees is 165°. To change the result from the unit radian to the unit degree multiply the angle by 180° / $\pi$ and obtain 165°.
Our results above contain fractions of pi for the results in radian, and are exact values otherwise. If you compute arccos(-(sqrt6 + sqrt2)/4), and any other angle, using the calculator below, then the value will be rounded to ten decimal places.
To obtain the angle in degrees insert -(sqrt6 + sqrt2)/4 as decimal in the field labelled “x”. However, if you want to be given the angle adjacent to -(sqrt6 + sqrt2)/4 in radians, then you must press the swap units button.
Calculate arccos x
The identities of arccosine -(sqrt6 + sqrt2)/4 are as follows: arccos(-(sqrt6 + sqrt2)/4) =
- $\frac{\pi}{2}$ – arcsin(-(sqrt6 + sqrt2)/4) ⇔ 90°- arcsin(-(sqrt6 + sqrt2)/4)
- ${\pi}$ – arccos((sqrt6 + sqrt2)/4) ⇔ 180° – arcos((sqrt6 + sqrt2)/4)
- arcsec(1/-(sqrt6 + sqrt2)/4)
- $2arctan(\frac{\sqrt{1-(-(\sqrt{6}+\sqrt{2})/4)^{2}}}{1+(-(\sqrt{6}+\sqrt{2})/4)})$
The infinite series of arccos -(sqrt6 + sqrt2)/4 is: $\frac{\pi}{2}$ – $\sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^{2}(2n+1)}(-(\sqrt{6}+\sqrt{2})/4)^{2n+1}$.
Next, we discuss the derivative of arccos x for x = -(sqrt6 + sqrt2)/4. In the following paragraph you can additionally learn what the search calculations form in the sidebar is used for.
Derivative of arccos -(sqrt6 + sqrt2)/4
The derivative of arccos -(sqrt6 + sqrt2)/4 is particularly useful to calculate the inverse cosine -(sqrt6 + sqrt2)/4 as an integral.
The formula for x is (arccos x)’ = – $\frac{1}{\sqrt{1-x^{2}}}$, x ≠ -1,1, so for x = -(sqrt6 + sqrt2)/4 the derivative equals -3.8637033052.
Using the arccos -(sqrt6 + sqrt2)/4 derivative, we can calculate the angle as a definite integral:
arccos -(sqrt6 + sqrt2)/4 = $\int_{-(\sqrt{6}+\sqrt{2})/4}^{1}\frac{1}{\sqrt{1-z^{2}}}dz$.
The relationship of arccos of -(sqrt6 + sqrt2)/4 and the trigonometric functions sin, cos and tan is:
- sin(arccosine(-(sqrt6 + sqrt2)/4)) =$\sqrt{1-(-(\sqrt{6}+\sqrt{2})/4)^{2}}$
- cos(arccosine(-(sqrt6 + sqrt2)/4)) = -(sqrt6 + sqrt2)/4
- tan(arccosine(-(sqrt6 + sqrt2)/4)) = $\frac{\sqrt{1-(-(\sqrt{6}+\sqrt{2})/4)^{2}}}{-(\sqrt{6}+\sqrt{2})/4}$
Note that you can locate many terms including the arccosine(-(sqrt6 + sqrt2)/4) value using the search form. On mobile devices you can find it by scrolling down. Enter, for instance, arccos-(sqrt6 + sqrt2)/4 angle.
Using the aforementioned form in the same way, you can also look up terms including derivative of inverse cosine -(sqrt6 + sqrt2)/4, inverse cosine -(sqrt6 + sqrt2)/4, and derivative of arccos -(sqrt6 + sqrt2)/4, just to name a few.
In the next part of this article we discuss the trigonometric significance of arccosine -(sqrt6 + sqrt2)/4, and there we also explain the difference between the inverse and the reciprocal of cos -(sqrt6 + sqrt2)/4.
What is arccos -(sqrt6 + sqrt2)/4?
In a triangle which has one angle of 90 degrees, the cosine of the angle α is the ratio of the length of the adjacent side a to the length of the hypotenuse h: cos α = a/h.
In a circle with the radius r, the horizontal axis x, and the vertical axis y, α is the angle formed by the two sides x and r; r moving counterclockwise defines the positive angle.
As follows from the unit-circle definition on our homepage, assumed r = 1, in the intersection of the point (x,y) and the circle, x = cos α = -(sqrt6 + sqrt2)/4 / r = -(sqrt6 + sqrt2)/4. The angle whose cosine value equals -(sqrt6 + sqrt2)/4 is α.
In the interval [0, pi] or [0°, 180°], there is only one α whose arccosine value equals -(sqrt6 + sqrt2)/4. For that interval we define the function which determines the value of α as
From the definition of arccos(-(sqrt6 + sqrt2)/4) follows that the inverse function y-1 = cos(y) = -(sqrt6 + sqrt2)/4. Observe that the reciprocal function of cos(y),(cos(y))-1 is 1/cos(y).
Avoid misconceptions and remember (cos(y))-1 = 1/cos(y) ≠ cos-1(y) = arccos(-(sqrt6 + sqrt2)/4). And make sure to understand that the trigonometric function y=arccos(x) is defined on a restricted domain, where it evaluates to a single value only, called the principal value:
In order to be injective, also known as one-to-one function, y = arccos(x) if and only if cos y = x and 0 ≤ y ≤ pi. The domain of x is −1 ≤ x ≤ 1.
Conclusion
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The frequently asked questions in the context include what is arccos -(sqrt6 + sqrt2)/4 degrees and what is the inverse cosine -(sqrt6 + sqrt2)/4 for example; reading our content they are no-brainers.
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– Article written by Mark, last updated on February 4th, 2017