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Welcome to **arccsc -(sqrt6 + sqrt2)**, our post aboutthe arccosecant of -(sqrt6 + sqrt2).

For the inverse trigonometric function of arccsc -(sqrt6 + sqrt2) we usually employ the abbreviation *arccsc* and write it as arccsc -(sqrt6 + sqrt2) or arccsc(-(sqrt6 + sqrt2)).

If you have been looking for *what is arccsc -(sqrt6 + sqrt2)*, either in degrees or radians, or if you have been wondering about the inverse of csc -(sqrt6 + sqrt2), then you are right here, too.

In this post you can find the angle arccosecant of -(sqrt6 + sqrt2), along with identities.

Read on to learn all about the arccsc of -(sqrt6 + sqrt2), and note that the term -sqrt6-sqrt2 is approximately -3.8637033 as a decimal number.

## Arccsc of -(sqrt6 + sqrt2)

If you want to know *what is arccsc -(sqrt6 + sqrt2)* in terms of trigonometry, check out the explanations in the last paragraph; ahead in this section is the value of arccosecant(-(sqrt6 + sqrt2)):

arccosecant -(sqrt6 + sqrt2) = -pi/12 rad = -15 °

arccosecant of -(sqrt6 + sqrt2) = -pi/12 radians = -15 degrees

The arccsc of -(sqrt6 + sqrt2) is -pi/12 radians, and the value in degrees is -15°. To change the result from the unit radian to the unit degree multiply the angle by 180° / $\pi$ and obtain -15°.

Our results above contain fractions of pi for the results in radian, and are exact values otherwise. If you compute arccsc(-(sqrt6 + sqrt2)), 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) as decimal in the field labelled “x”. However, if you want to be given the angle of csc -(sqrt6 + sqrt2) in radians, then you must press the swap units button.

### Calculate arccsc x

The identities of arccosecant -(sqrt6 + sqrt2) are as follows: arccsc(-(sqrt6 + sqrt2)) =

- $\frac{\pi}{2}$ – arcsec(-(sqrt6 + sqrt2)) ⇔ 90°- arcsec(-(sqrt6 + sqrt2))
- -arccsc((sqrt6 + sqrt2))
- arcsin(1/-(sqrt6 + sqrt2))

The infinite series of arccsc -(sqrt6 + sqrt2) is: $\sum_{n=0}^{\infty}\frac{\binom{2n}{n}(-(\sqrt{6}+\sqrt{2}))^{-(2n+1)}}{4^{n}(2n+1)}$.

Next, we discuss the derivative of arccsc -(sqrt6 + sqrt2) for -(sqrt6 + sqrt2) = -(sqrt6 + sqrt2). In the following paragraph you can additionally learn what the search calculations form in the sidebar is used for.

## Derivative of arccsc -(sqrt6 + sqrt2)

The derivative of arccsc -(sqrt6 + sqrt2) is particularly useful to calculate the inverse arccsc -(sqrt6 + sqrt2) as an integral.

The formula for x is (arccsc x)’ = -$\frac{1}{|x|\sqrt{x^{2}-1}}$, |x| > 1, so for x = -(sqrt6 + sqrt2) the derivative equals -0.0693503541.

Using the arccsc -(sqrt6 + sqrt2) derivative, we can calculate the angle as a definite integral:

arccsc -(sqrt6 + sqrt2) = $\int_{-(\sqrt{6}+\sqrt{2})}^{\infty}\frac{1}{z\sqrt{z^{2}-1}}dz$.

The relationship of arccsc of -(sqrt6 + sqrt2) and the trigonometric functions sin, cos and tan is:

- sin(arccosecant(-(sqrt6 + sqrt2))) = $\frac{1}{(-(\sqrt{6}+\sqrt{2}))}$
- cos(arccosecant(-(sqrt6 + sqrt2))) = $\frac{\sqrt{(-(\sqrt{6}+\sqrt{2}))^{2}-1}}{-(\sqrt{6}+\sqrt{2})}$
- tan(arccosecant(-(sqrt6 + sqrt2))) = $\frac{1}{\sqrt{(-(\sqrt{6}+\sqrt{2}))^{2}-1}}$

Note that you can locate many terms including the arccosecant(-(sqrt6 + sqrt2)) value using the search form. On mobile devices you can find it by scrolling down. Enter, for instance, arccsc-(sqrt6 + sqrt2) angle.

Using the aforementioned form in the same way, you can also look up terms including derivative of inverse arccsc -(sqrt6 + sqrt2), inverse arccsc -(sqrt6 + sqrt2), and derivative of arccsc -(sqrt6 + sqrt2), just to name a few.

In the next part of this article we discuss the trigonometric significance of arccosecant -(sqrt6 + sqrt2), and there we also explain the difference between the inverse and the reciprocal of csc -(sqrt6 + sqrt2).

## What is arccsc -(sqrt6 + sqrt2)?

In a triangle which has one angle of 90 degrees, the sine of the angle α is the ratio of the length of the opposite side o to the length of the hypotenuse h: sin α = o/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, sin α = y / r = y, and csc α = 1 / y = -(sqrt6 + sqrt2). The angle whose arccsc value equals -(sqrt6 + sqrt2) is α.

In the interval [-pi/2, 0[ ∪ ]0, pi/2] or [-90°, 0[ ∪ ]0, 90°], there is only one α whose sine value equals -(sqrt6 + sqrt2). For that interval we define the function which determines the value of α as

From the definition of arccsc(-(sqrt6 + sqrt2)) follows that the *inverse* function y^{-1} = csc(y) = -(sqrt6 + sqrt2). Observe that the *reciprocal* function of csc(y),(csc(y))^{-1} is 1/csc(y) = sin(y).

Avoid misconceptions and remember (csc(y))^{-1} = 1/csc(y) ≠ csc^{-1}(y) = arccsc(-(sqrt6 + sqrt2)). And make sure to understand that the trigonometric function y=arccsc(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 = arccsc(x) if and only if csc y = x and -pi/2 ≤ y < 0 or csc y = x and 0 < y ≤ pi/2. The domain of x is x ≤ −1 or 1 ≤ x.

## Conclusion

The frequently asked questions in the context include *what is arccsc -(sqrt6 + sqrt2) degrees* and *what is the inverse arccsc -(sqrt6 + sqrt2)* for example; reading our content they are no-brainers.

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– Article written by Mark, last updated on February 4th, 2017