Inverse Trig Evaluation & Foundations of Trigonometric Identities

Evaluating Expressions With Inverse Trig Functions

  • Core idea: an inverse‐trig term (e.g. \cos^{-1}(\tfrac23)) represents an angle.

  • Strategy for exact evaluations

    1. Let a symbol (e.g. \theta) equal the inverse expression.

    2. Rewrite using the direct function.

    3. Sketch a right triangle that matches the ratio.

    4. Use the Pythagorean Theorem to find the missing side.

    5. Evaluate the requested trig ratio from the picture.

Example 1 – \tan\bigl(\cos^{-1}(\tfrac23)\bigr)

  • Let \theta=\cos^{-1}(\tfrac23) \;\Longrightarrow\; \cos\theta=\tfrac23.

  • Build a triangle (adjacent = 2, hypotenuse = 3).
    a^2+2^2=3^2 \;\Rightarrow\; a=\sqrt5.

  • Desired ratio: \tan\theta=\tfrac{\text{opposite}}{\text{adjacent}}=\tfrac{\sqrt5}{2}.

Example 2 – \cos\bigl(\sin^{-1}(\tfrac35)\bigr)

  • u=\sin^{-1}(\tfrac35) \Rightarrow \sin u=\tfrac35.

  • Triangle: opposite = 3, hypotenuse = 5, adjacent = 4 (3-4-5 triple).

  • \cos u=\tfrac45.

Example 3 – Algebraic form \sin\bigl(\cos^{-1}(3x)\bigr)

  • \alpha=\cos^{-1}(3x)\Rightarrow\cos\alpha=3x/1.

  • Triangle: adjacent = 3x, hypotenuse = 1, opposite = \sqrt{1-9x^2}.

  • \sin\alpha=\tfrac{\sqrt{1-9x^2}}{1}=\sqrt{1-9x^2}.

What Is an Identity?

  • Equation true for every permissible value of the variable, e.g. 3x+4x\equiv7x.

  • Contrast with conditional equations (true only for specific values, e.g. 2x+3=7 when x=2).

Families of Identities Already Used

  • Reciprocal
    \sin\theta=\tfrac1{\csc\theta},\; \cos\theta=\tfrac1{\sec\theta},\; \tan\theta=\tfrac1{\cot\theta} (and reciprocals reversed).

  • Pythagorean
    \sin^2\theta+\cos^2\theta=1, \quad 1+\tan^2\theta=\sec^2\theta, \quad 1+\cot^2\theta=\csc^2\theta.

  • Negative-angle (odd/even)
    \sin(-\theta)=-\sin\theta,\; \cos(-\theta)=\cos\theta,\; \tan(-\theta)=-\tan\theta.

  • Cofunction: cofunctions of complementary angles are equal
    \sin(90^\circ-\theta)=\cos\theta,\; \tan(90^\circ-\theta)=\cot\theta,\; \sec(90^\circ-\theta)=\csc\theta.

Finding Missing Trig Values From One Given Value

Example: \cot x=-3,\; \cos x>0.

  1. Quadrant analysis: \cot<0, \cos>0 \Rightarrow Quadrant IV.

  2. Immediate reciprocal: \tan x=-\tfrac13.

  3. Pythagorean link:
    1+\cot^2x=\csc^2x \;\Rightarrow\; 1+9=10 \Rightarrow \csc x=-\sqrt{10} (negative in QIV).

  4. \sin x=-1/\csc x=-\tfrac1{\sqrt{10}}=-\tfrac{\sqrt{10}}{10}.

  5. Use \cot x=\tfrac{\cos x}{\sin x} to solve for \cos x:
    \cos x=\cot x\,\sin x=(-3)\Bigl(-\tfrac{\sqrt{10}}{10}\Bigr)=\tfrac{3\sqrt{10}}{10}>0.

  6. \sec x=\tfrac1{\cos x}=\tfrac{10}{3\sqrt{10}}=\tfrac{\sqrt{10}}3.

Simplifying Expressions – Sample Techniques

  • Factor common trig terms: \sin x(\cos^2x-1)\to\sin x(-\sin^2x)=-\sin^3x.

  • Convert to \sin,\cos then combine fractions:
    \sin t+\cot t\,\cos t=\sin t+\tfrac{\cos^2t}{\sin t}=\tfrac{\sin^2t+\cos^2t}{\sin t}=\tfrac1{\sin t}=\csc t.

  • Factor as difference of squares:
    \tfrac{\sin^2x-1}{\sin x-1}=\tfrac{(\sin x+1)(\sin x-1)}{\sin x-1}=\sin x+1.

Proving Identities – General Guidelines

  • Work one side only, preferably the more complicated.

  • Legal moves: replace with established identities + ordinary algebra (factor, expand, common denominator, cancel common factors, conjugates, etc.).

  • Document steps mentally (e.g. “reciprocal identity”, “Pythagorean substitution”, “algebraic cancellation”).

  • Typical rescue moves
    • Convert everything to \sin,\cos.
    • Multiply by a conjugate 1\pm\cos\theta or 1\pm\sin\theta to force \sin^2+\cos^2=1.
    • Separate a single fraction’s numerator: \tfrac{A+B}{D}=\tfrac AD+\tfrac BD (never split denominators!).

Worked Proofs

  1. \sec\theta\,\cot\theta\stackrel{?}{=}\csc\theta
    \sec\theta\,\cot\theta=\tfrac1{\cos\theta}\,\tfrac{\cos\theta}{\sin\theta}=\tfrac1{\sin\theta}=\csc\theta.

  2. \cos u-\sec u\,\sin^2u\equiv\cos^2u
    \cos u-\tfrac1{\cos u}\,\sin^2u=\tfrac{\cos^2u-\sin^2u}{\cos u}=\tfrac{1-\sin^2u-\sin^2u}{\cos u}=\tfrac{1-2\sin^2u}{\cos u} (continue algebra → result \cos^2u).

  3. \cos x-\sec x\;\stackrel{?}{=}-\sin x\,\tan x
    Convert \sec x=\tfrac1{\cos x}, get common denominator, identify \sin^2x=1-\cos^2x → simplifies to RHS.

  4. Conjugate trick:
    Prove \tfrac{\sin\theta}{1-\cos\theta}=\tfrac{1+\cos\theta}{\sin\theta}.

    • Multiply by \tfrac{1+\cos\theta}{1+\cos\theta} → denominator 1-\cos^2\theta=\sin^2\theta; cancel \sin\theta.

  5. Polynomial-looking identity:
    3\cos^4\theta+6\sin^2\theta\equiv3+3\sin^4\theta
    Replace \cos^2\theta by 1-\sin^2\theta, square, distribute, like terms cancel → RHS.

  6. Difference of 4th powers:
    \sec^4\theta-\tan^4\theta=(\sec^2\theta-\tan^2\theta)(\sec^2\theta+\tan^2\theta)=1\,(\sec^2\theta+\tan^2\theta).

Practical Advice & Mindset

  • Do something – staring rarely yields insight; even a wrong start sparks new neural paths.

  • When you see 1\pm \text{trig} in a denominator, think conjugate.

  • Factor patterns (difference of squares, square of binomial) appear constantly – treat trig symbols like ordinary algebraic letters.

  • Never operate on both sides simultaneously unless you meet in the middle and show the same expression.

  • Write steps clearly so a peer “not quite as smart as you” can follow.