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Arctg -2+sqrt3

Welcome to arctg -2+sqrt3, our post aboutthe arctangent of -2+sqrt3.

For the inverse trigonometric function of tangent -2+sqrt3 we usually employ the abbreviation arctg and write it as arctg -2+sqrt3 or arctg(-2+sqrt3).

If you have been looking for what is arctg -2+sqrt3, either in degrees or radians, or if you have been wondering about the inverse of tg -2+sqrt3, then you are right here, too.

In this post you can find the angle arctangent of -2+sqrt3, along with identities.

Read on to learn all about the arctg of -2+sqrt3, and note that the term -2+sqrt3 is approximately -0.26794919 as a decimal number.

Arctg of -2+sqrt3

If you want to know what is arctg -2+sqrt3 in terms of trigonometry, check out the explanations in the last paragraph; ahead in this section is the value of arctangent(-2+sqrt3):

arctg -2+sqrt3 = -π/12 rad = -15°
arctangent -2+sqrt3 = -π/12 rad = -15 °
arctangent of -2+sqrt3 = -π/12 radians = -15 degrees

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The arctg of -2+sqrt3 is -π/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 π for the results in radian, and are exact values otherwise. If you compute arctg(-2+sqrt3), 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 -2+sqrt3 as decimal in the field labelled “x”. However, if you want to be given the angle of tg -2+sqrt3 in radians, then you must press the swap units button.

Calculate arctg x

A Really Cool Arctangent Calculator and Useful Information! Please ReTweet. Click To TweetApart from the inverse of tg -2+sqrt3, similar trigonometric calculations include:

The identities of arctangent -2+sqrt3 are as follows: arctg(-2+sqrt3) =

  • $\frac{\pi}{2}$ – arcctg(-2+sqrt3) ⇔ 90°- arcctg(-2+sqrt3)
  • -arctg(2+√3)
  • arcctg(1/-2+sqrt3) – $\pi $ ⇔ arcctg(1/-2+sqrt3) – 180°
  • $arcsin(\frac{-2+\sqrt{3}}{\sqrt{(-2+\sqrt{3})^{2}+1}})$
  • $2arctg(\frac{-2+\sqrt{3}}{1+\sqrt{(-2+\sqrt{3})^{2}+1}})$

The infinite series of arctg -2+sqrt3 is: $\sum_{n=0}^{\infty} \frac{(-1)^{n}(-2+\sqrt{3})^{2n+1}}{(2n+1)}$.

Next, we discuss the derivative of arctg -2+sqrt3 for x = -2+sqrt3. In the following paragraph you can additionally learn what the search calculations form in the sidebar is used for.

Derivative of arctg -2+sqrt3

The derivative of arctg -2+sqrt3 is particularly useful to calculate the inverse tangent -2+sqrt3 as an integral.

The formula for x is (arctg x)’ = $\frac{1}{1 + x^{2}}$, x ≠ -1,1, so for x = -2+sqrt3 the derivative equals 0.9330127019.

Using the arctg -2+sqrt3 derivative, we can calculate the angle as a definite integral:

arctg -2+sqrt3 = $\int_{0}^{-2+\sqrt{3}}\frac{1}{{1+z^{2}}}dz$.

The relationship of arctg of -2+sqrt3 and the trigonometric functions sin, cos and tg is:

  • sin(arctangent(-2+sqrt3)) = $\frac{-2+\sqrt{3}}{\sqrt{1 + (-2+\sqrt{3})^{2}}}$
  • cos(arctangent(-2+sqrt3)) = $\frac{1}{\sqrt{1 + (-2+\sqrt{3})^{2}}}$
  • tg(arctangent(-2+sqrt3)) = -2+sqrt3

Note that you can locate many terms including the arctangent(-2+sqrt3) value using the search form. On mobile devices you can find it by scrolling down. Enter, for instance, arctg-2+sqrt3 angle.

Using the aforementioned form in the same way, you can also look up terms including derivative of inverse tangent -2+sqrt3, inverse tangent -2+sqrt3, and derivative of arctg -2+sqrt3, just to name a few.

In the next part of this article we discuss the trigonometric significance of arctangent -2+sqrt3, and there we also explain the difference between the inverse and the reciprocal of tg -2+sqrt3.

What is arctg -2+sqrt3?

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, and 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, sin α = y / r = y, cos α = x / r = x, and tg α = y / x = -2+sqrt3. The angle whose tangent value equals -2+sqrt3 is α.

In the interval ]-π/2, π/2[ or ]-90°, 90°[, there is only one α whose arctangent value equals -2+sqrt3. For that interval we define the function which determines the value of α as

y = arctg(-2+sqrt3).

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From the definition of arctg(-2+sqrt3) follows that the inverse function y-1 = tg(y) = -2+sqrt3. Observe that the reciprocal function of tg(y),(tg(y))-1 is 1/tg(y).

Avoid misconceptions and remember (tg(y))-1 = 1/tg(y) ≠ tg-1(y) = arctg(-2+sqrt3). And make sure to understand that the trigonometric function y=arctg(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 = arctg(x) if and only if tg y = x and -π/2 < y < π/2. The domain of x is $\mathbb{R}$.

Conclusion

Arctg -2+sqrt3The frequently asked questions in the context include what is arctg -2+sqrt3 degrees and what is the inverse tangent -2+sqrt3 for example; reading our content they are no-brainers.

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– Article written by Mark, last updated on February 6th, 2017

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