Math Problem Statement

Rewrite the expressions involving inverse trigonometric functions into algebraic expressions in x.

Solution

Let’s solve the given problems:

35. cos(sin1(x))\cos(\sin^{-1}(x))

We know that if sin1(x)=θ\sin^{-1}(x) = \theta, then sin(θ)=x\sin(\theta) = x. Using the Pythagorean identity:

cos2(θ)+sin2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1

cos2(θ)=1sin2(θ)\cos^2(\theta) = 1 - \sin^2(\theta)

cos(θ)=1x2\cos(\theta) = \sqrt{1 - x^2}

Thus:

cos(sin1(x))=1x2\cos(\sin^{-1}(x)) = \sqrt{1 - x^2}


36. sin(tan1(x))\sin(\tan^{-1}(x))

If tan1(x)=θ\tan^{-1}(x) = \theta, then tan(θ)=x\tan(\theta) = x. From the right triangle definition, the sides are:

  • Opposite = xx
  • Adjacent = 11
  • Hypotenuse = x2+1\sqrt{x^2 + 1}

sin(θ)=OppositeHypotenuse=xx2+1\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{x}{\sqrt{x^2 + 1}}

Thus:

sin(tan1(x))=xx2+1\sin(\tan^{-1}(x)) = \frac{x}{\sqrt{x^2 + 1}}


37. tan(sin1(x))\tan(\sin^{-1}(x))

If sin1(x)=θ\sin^{-1}(x) = \theta, then sin(θ)=x\sin(\theta) = x. From the Pythagorean identity:

  • Opposite = xx
  • Hypotenuse = 11
  • Adjacent = 1x2\sqrt{1 - x^2}

tan(θ)=OppositeAdjacent=x1x2\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{x}{\sqrt{1 - x^2}}

Thus:

tan(sin1(x))=x1x2\tan(\sin^{-1}(x)) = \frac{x}{\sqrt{1 - x^2}}


38. cos(tan1(x))\cos(\tan^{-1}(x))

If tan1(x)=θ\tan^{-1}(x) = \theta, then tan(θ)=x\tan(\theta) = x. From the triangle:

  • Opposite = xx
  • Adjacent = 11
  • Hypotenuse = x2+1\sqrt{x^2 + 1}

cos(θ)=AdjacentHypotenuse=1x2+1\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}}

Thus:

cos(tan1(x))=1x2+1\cos(\tan^{-1}(x)) = \frac{1}{\sqrt{x^2 + 1}}


Summary:

  1. cos(sin1(x))=1x2\cos(\sin^{-1}(x)) = \sqrt{1 - x^2}
  2. sin(tan1(x))=xx2+1\sin(\tan^{-1}(x)) = \frac{x}{\sqrt{x^2 + 1}}
  3. tan(sin1(x))=x1x2\tan(\sin^{-1}(x)) = \frac{x}{\sqrt{1 - x^2}}
  4. cos(tan1(x))=1x2+1\cos(\tan^{-1}(x)) = \frac{1}{\sqrt{x^2 + 1}}

Let me know if you'd like further explanation!


Relative Questions:

  1. How do trigonometric identities relate to inverse functions?
  2. What is the geometric interpretation of the inverse trigonometric functions?
  3. How can these results be used in applied problems involving angles?
  4. What happens if xx exceeds the domain of certain inverse trigonometric functions?
  5. How can we generalize these results to different coordinate systems?

Tip: Always draw a right triangle when dealing with inverse trigonometric expressions to clearly visualize the relationships between sides and angles.

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Math Problem Analysis

Mathematical Concepts

Trigonometric Functions
Inverse Trigonometric Functions
Algebraic Identities
Pythagorean Identity

Formulas

cos^2(θ) + sin^2(θ) = 1
sin(θ) = opposite / hypotenuse
tan(θ) = opposite / adjacent
cos(θ) = adjacent / hypotenuse

Theorems

Pythagorean Theorem
Definitions of Trigonometric Functions in Right Triangles

Suitable Grade Level

Grades 11-12