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Inverse trigonometric functions

arcsin, arccos, arctan (which angle has this ratio) and their derivatives.

sin is not one-to-one, so its inverse is defined on a restricted range: arcsin returns an angle in [−π/2, π/2], arccos in [0, π], arctan in (−π/2, π/2). Picture it: reflect the sine curve across y = x and keep one branch. Think it: their derivatives are algebraic (1/√(1 − x²), 1/(1 + x²)), which is why they appear as antiderivatives of ordinary fractions.

ምሳሌ · derivative of arctan(x)

Differentiate atan(x)

\frac{d}{dx}\left[\operatorname{atan}{\left(x \right)}\right]

እርምጃ በደረጃ

  1. \frac{d}{dx}\left[\operatorname{atan}{\left(x \right)}\right]

    Start from the derivative to compute.

  2. \frac{1}{x^{2} + 1}

    (arctan u)′ = 1/(1+u²).

መልሱን አሳይ
f'(x) = \frac{1}{x^{2} + 1}

አሁን ችግርን ይምረጡ፣ ወይም ጻፉ ወይም በራሳችሁ ቅርጽ አቅርቡ። እርምጃው ሁሉ፣ ምስል፣ መልሱ እስከምትጠይቅ ድረስ የተደበቀ ነው።

የራስዎን ስራዎች ይያዙ

ነጻ የሆኑት መተግበሪያዎች በየክፍል ውስጥ ማስታወሻዎችን ያካትታሉ፣ የተጠናቀቁትን ነገሮች መዝገብ፣ የተፈቱ ችግሮችን በአንድ ቦታ፣ ስለዚህ ገጽ መጠየቅ የሚችሉትን ረዳት ያካትታሉ፡፡ የቁጥር ትምህርት ለሁሉም ሰው ክፍት ነው፣ ተቀባይነት ያለው ወይም የለም፤

ምዝገባ መግባቱ

ፊደላት

ለሙሉ መግለጫ፣ ምስል፣ እና በውስጡ ያለውን ፊደል ትርጓሜ ለማየት ማንኛውንም ምልክት ይጫኑ።

እንዴት: Inverse trigonometric functions

  1. Start from the derivative to compute.
  2. (arctan u)′ = 1/(1+u²).

ጥያቄዎች

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

በ Trigonometry