Math127 Inverse Trig Functions continued

Inverse Trigonometric Functions

• Overview: Inverse trigonometric functions and their domains are crucial in trigonometry.

Domain and Range

• Importance of understanding the domain and range for inverse functions.

• Values in trigonometry must adhere to the designated domain.

Secant, Cosecant, and Cotangent

• Discussed functions beyond sine, cosine, and tangent.

Secant Inverse of One

• To find secant inverse of 1: secant = 1/(cosine)

• Secant inverse of 1 relates to cosine = 1/1.

• Height (a) is calculated as:

  • a = (\sqrt{1^2 - 1^2} = 0)

• Therefore, secant inverse of 1 = 0 degrees.

Cosecant Inverse of Two

• Cosecant inverse of 2 means: 2/1

• Converts to sine inverse as 1/2 (flipping values).

• Triangle parameters: height = 1, hypotenuse = 2.

• Side (b) calculated as:

  • b = (\sqrt{2^2 - 1^2} = \sqrt{3})

• Result: cosecant inverse of 2 = (\frac{\pi}{6}) or 30 degrees (first quadrant).

Cotangent Inverse of Negative (\sqrt{3}) over 1

• Cotangent inverse of -(\sqrt{3}): equivalent to tan inverse of -(\frac{1}{\sqrt{3}}).

• Triangle sides: 1 and (\sqrt{3}).

• Hypotenuse = (\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2).

• This gives: cotangent inverse of -(\sqrt{3}) = (\frac{5\pi}{6}) (second quadrant).

Example Problem: Angle of a Triangle with Cosine (\frac{3}{5})

• To determine triangle angle with cosine: (\frac{3}{5}).

• Find side 'b' using the Pythagorean theorem:

  • (b = \sqrt{25 - 9} = 4)

• Consequently:

  • Sine = (\frac{4}{5})

• Since cosine is positive, angle resides in the first quadrant.

Example Problem: Cosine of Sine Inverse of Negative (\frac{1}{\sqrt{16}})

• Triangle value: sine = (\frac{1}{\sqrt{6}}).

• Since sine is negative, angle is in the fourth quadrant.

• Side 'a' is calculated as:

  • (\sqrt{6 - 1} = \sqrt{5})

• Cosine = (\frac{\sqrt{5}}{\sqrt{6}})

  • Rationalized to (\frac{\sqrt{30}}{6}).

Example Problem: Tangent as Sine Inverse of Negative (\frac{3}{4})

• Derived cosine: (\frac{3}{4}) and determined quadrant based on sine negativity (2nd quadrant).

• Evaluating 'b':

  • (b = \sqrt{4^2 - 3^2} = \sqrt{16 - 9} = \sqrt{7}).

• Thus, tangent = (\frac{\sqrt{7}}{3}) in the quadrants considered.

Example Problem: Cosecant with Tangent

• Tangent = (\frac{1}{3}), determined quadrant based on negative value (4th quadrant).

• Hypotenuse calculation: (\sqrt{3^2 + 1^2} = \sqrt{10}).

• Sine and cosecant findings:

  • Sine = (\frac{1}{\sqrt{10}}), hence cosecant = (\sqrt{10}) (must be negative due to quadrant position).

Example Problem: Tangent Inverse of (\frac{7}{24})

• Derived triangle sides yield: sine = (\frac{7}{25}).

• Evaluated quadrant placement on tangent positivity (1st quadrant).

• Summary of positive angles and sine from confirmed placements.

Inverse Trigonometric FunctionsOverview: Understanding inverse trigonometric functions and their domains is crucial in trigonometry.

• Domain and Range: Vital for inverse functions; values must adhere to the designated domain.

• Secant: Secant inverse of 1 equates to 0 degrees (a = 0).

• Cosecant: Cosecant inverse of 2 gives 30 degrees (b = (\sqrt{3})).

• Cotangent: Cotangent inverse of -(\sqrt{3}) results in 150 degrees.

• Example Problems:

1. Cosine of (\frac{3}{5}) leads to angle in first quadrant.

2. Sine inverse of -(\frac{1}{\sqrt{16}}) yields angle in fourth quadrant.

3. Tangent inverse of -(\frac{3}{4}) indicates second quadrant position.

4. Cosecant related to tangent dimensions in fourth quadrant.

5. Tangent inverse of (\frac{7}{24}) reveals first quadrant placement and angle validation.