Inverse Trig

Definition of \arctan x

• Step 1: Let (y = \arctan x). This means "y is the angle whose tangent is x."* Step 2: By definition: ( \tan y = x)* Step 3: Domain for arctan: (y \in (-\pi/2, \pi/2)) (first and fourth quadrants)* Step 4: Why it matters: The output is always an angle; this lets us use triangles to simplify expressions like \sin(\arctan x) or \cos(\arctan x) in integrals. Example: \arctan 1 = \pi/4 because \tan(\pi/4) = 1

Definition of \arcsin x

• Step 1: Let (y = \arcsin x). This means "y is the angle whose sine is x."* Step 2: By definition: ( \sin y = x)* Step 3: Domain for arcsin: (y \in [-\pi/2, \pi/2]) (first and fourth quadrants)* Step 4: Why it matters: Output is always an angle; allows simplifications like \cos(\arcsin x) = \sqrt{1-x^2} Example: \arcsin(1/2) = \pi/6 because \sin(\pi/6) = 1/2

Definition of \arccos x

• Step 1: Let (y = \arccos x). This means "y is the angle whose cosine is x."* Step 2: By definition: ( \cos y = x)* Step 3: Domain for arccos: (y \in [0, \pi]) (first and second quadrants)* Step 4: Why it matters: Output is always an angle; allows simplifications like \sin(\arccos x) = \sqrt{1-x^2} Example: \arccos(1/2) = \pi/3 because \cos(\pi/3) = 1/2

Simplifying \sin(\arctan x) using a triangle

• Step 1: Let (y = \arctan x) \rightarrow (\tan y = x/1)* Step 2: Draw a right triangle: opposite = x, adjacent = 1* Step 3: Hypotenuse = ( \sqrt{1+x^2}) (Pythagoras)* Step 4: ( \sin y = \text{opposite}/\text{hypotenuse} = x/\sqrt{1+x^2})* Step 5: ( \cos y = \text{adjacent}/\text{hypotenuse} = 1/\sqrt{1+x^2}) Tip: Always draw a triangle for inverse trig expressions.

Simplifying \cos(2 \arctan x)

• Step 1: Let (y = \arctan x) \rightarrow (\tan y = x)* Step 2: Use double angle formula: ( \cos 2y = \cos^2 y - \sin^2 y)* Step 3: Substitute triangle ratios:* ( \cos^2 y = (1/\sqrt{1+x^2})^2 = 1/(1+x^2))* ( \sin^2 y = (x/\sqrt{1+x^2})^2 = x^2/(1+x^2))* Step 4: Subtract: ( \cos 2y = 1/(1+x^2) - x^2/(1+x^2) = (1-x^2)/(1+x^2))

Derivative of \arcsin x

• \frac{d}{dx}(\arcsin x) = 1/\sqrt{1-x^2}* Baby steps:* Think: \arcsin x \u2192 "angle whose sine is x"* Use implicit differentiation: \sin y = x \implies \cos y , dy/dx = 1 \implies dy/dx = 1/\cos y = 1/\sqrt{1-x^2}

Derivative of \arccos x

• \frac{d}{dx}(\arccos x) = -1/\sqrt{1-x^2}* Baby steps:* Start: ( \cos y = x)* Implicit diff: ( -\sin y , dy/dx = 1 \implies dy/dx = -1/\sin y = -1/\sqrt{1-x^2})

Derivative of \arctan x

• \frac{d}{dx}(\arctan x) = 1/(1+x^2)* Baby steps:* Start: ( \tan y = x)* Implicit diff: ( \sec^2 y , dy/dx = 1 \implies dy/dx = 1/\sec^2 y = 1/(1+\tan^2 y) = 1/(1+x^2))

Definition of \arctan x

• Step 1: Understanding the Inverse Function

  • Let ( y = \arctan x ). This statement means "y is the angle whose tangent is x."

  • Essentially, we are looking for an angle y in the restricted domain whose tangent value is x.

• Step 2: The Core Definition

  • By definition of the inverse tangent: ( \tan y = x ). This rephrases the relationship in terms of the original tangent function.

• Step 3: Domain and Range for \arctan x

  • The restricted domain for tangent, which becomes the range for arc-tangent, is ( y \in (-\pi/2, \pi/2) ).

  • This interval represents angles exclusively in the first and fourth quadrants (excluding the vertical asymptotes at \pm \pi/2).

  • The output, y, is always an angle.

• Step 4: Practical Significance and Simplification

  • Knowing that the output is always an angle allows us to use right triangles to simplify complex expressions involving inverse trigonometric functions.

  • For example, you can simplify expressions like \sin(\arctan x) or \cos(\arctan x) which frequently appear in integral problems.

  • Example: \arctan 1 = \pi/4 because \tan(\pi/4) = 1. Here, the input 1 (a ratio) gives an output \pi/4 (an angle).

Integral: \int \frac{dx}{a^2+x^2}

• Step 1: Recognize: derivative of \arctan x formula* Step 2: ( \int \frac{dx}{a^2+x^2} = (1/a) \arctan(x/a) + C)* Step 3: Always check for constant ",a" factor