Inverse Trig Evaluation & Foundations of Trigonometric Identities

Evaluating Expressions With Inverse Trig Functions

  • Core idea: an inverse‐trig term (e.g. cos⁡−1(23)\cos^{-1}(\tfrac23)) represents an angle.

  • Strategy for exact evaluations

    1. Let a symbol (e.g. θ\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⁡(cos⁡−1(23))\tan\bigl(\cos^{-1}(\tfrac23)\bigr)

  • Let θ=cos⁡−1(23)  ⟹  cos⁡θ=23.\theta=\cos^{-1}(\tfrac23) \;\Longrightarrow\; \cos\theta=\tfrac23.

  • Build a triangle (adjacent = 2, hypotenuse = 3).
    a2+22=32  ⇒  a=5.a^2+2^2=3^2 \;\Rightarrow\; a=\sqrt5.

  • Desired ratio: tan⁡θ=oppositeadjacent=52.\tan\theta=\tfrac{\text{opposite}}{\text{adjacent}}=\tfrac{\sqrt5}{2}.

Example 2 – cos⁡(sin⁡−1(35))\cos\bigl(\sin^{-1}(\tfrac35)\bigr)

  • u=sin⁡−1(35)⇒sin⁡u=35.u=\sin^{-1}(\tfrac35) \Rightarrow \sin u=\tfrac35.

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

  • cos⁡u=45.\cos u=\tfrac45.

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

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

  • Triangle: adjacent = 3x3x, hypotenuse = 1, opposite = 1−9x2\sqrt{1-9x^2}.

  • sin⁡α=1−9x21=1−9x2.\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≡7x.3x+4x\equiv7x.

  • Contrast with conditional equations (true only for specific values, e.g. 2x+3=72x+3=7 when x=2x=2).

Families of Identities Already Used

  • Reciprocal
    sin⁡θ=1csc⁡θ,  cos⁡θ=1sec⁡θ,  tan⁡θ=1cot⁡θ\sin\theta=\tfrac1{\csc\theta},\; \cos\theta=\tfrac1{\sec\theta},\; \tan\theta=\tfrac1{\cot\theta} (and reciprocals reversed).

  • Pythagorean
    sin⁡2θ+cos⁡2θ=1,1+tan⁡2θ=sec⁡2θ,1+cot⁡2θ=csc⁡2θ.\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⁡(−θ)=−sin⁡θ,  cos⁡(−θ)=cos⁡θ,  tan⁡(−θ)=−tan⁡θ.\sin(-\theta)=-\sin\theta,\; \cos(-\theta)=\cos\theta,\; \tan(-\theta)=-\tan\theta.

  • Cofunction: cofunctions of complementary angles are equal
    sin⁡(90∘−θ)=cos⁡θ,  tan⁡(90∘−θ)=cot⁡θ,  sec⁡(90∘−θ)=csc⁡θ.\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=−13.\tan x=-\tfrac13.

  3. Pythagorean link:
    1+cot⁡2x=csc⁡2x  ⇒  1+9=10⇒csc⁡x=−101+\cot^2x=\csc^2x \;\Rightarrow\; 1+9=10 \Rightarrow \csc x=-\sqrt{10} (negative in QIV).

  4. sin⁡x=−1/csc⁡x=−110=−1010.\sin x=-1/\csc x=-\tfrac1{\sqrt{10}}=-\tfrac{\sqrt{10}}{10}.

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

  6. sec⁡x=1cos⁡x=10310=103.\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)→sin⁡x(−sin⁡2x)=−sin⁡3x.\sin x(\cos^2x-1)\to\sin x(-\sin^2x)=-\sin^3x.

  • Convert to sin⁡,cos⁡\sin,\cos then combine fractions:
    sin⁡t+cot⁡t cos⁡t=sin⁡t+cos⁡2tsin⁡t=sin⁡2t+cos⁡2tsin⁡t=1sin⁡t=csc⁡t.\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:
    sin⁡2x−1sin⁡x−1=(sin⁡x+1)(sin⁡x−1)sin⁡x−1=sin⁡x+1.\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⁡\sin,\cos.
    • Multiply by a conjugate 1±cos⁡θ1\pm\cos\theta or 1±sin⁡θ1\pm\sin\theta to force sin⁡2+cos⁡2=1\sin^2+\cos^2=1.
    • Separate a single fraction’s numerator: A+BD=AD+BD\tfrac{A+B}{D}=\tfrac AD+\tfrac BD (never split denominators!).

Worked Proofs

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

  2. cos⁡u−sec⁡u sin⁡2u≡cos⁡2u\cos u-\sec u\,\sin^2u\equiv\cos^2u
    cos⁡u−1cos⁡u sin⁡2u=cos⁡2u−sin⁡2ucos⁡u=1−sin⁡2u−sin⁡2ucos⁡u=1−2sin⁡2ucos⁡u\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\cos^2u).

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

  4. Conjugate trick:
    Prove sin⁡θ1−cos⁡θ=1+cos⁡θsin⁡θ.\tfrac{\sin\theta}{1-\cos\theta}=\tfrac{1+\cos\theta}{\sin\theta}.

    • Multiply by 1+cos⁡θ1+cos⁡θ\tfrac{1+\cos\theta}{1+\cos\theta} → denominator 1−cos⁡2θ=sin⁡2θ1-\cos^2\theta=\sin^2\theta; cancel sin⁡θ\sin\theta.

  5. Polynomial-looking identity:
    3cos⁡4θ+6sin⁡2θ≡3+3sin⁡4θ3\cos^4\theta+6\sin^2\theta\equiv3+3\sin^4\theta
    Replace cos⁡2θ\cos^2\theta by 1−sin⁡2θ1-\sin^2\theta, square, distribute, like terms cancel → RHS.

  6. Difference of 4th powers:
    sec⁡4θ−tan⁡4θ=(sec⁡2θ−tan⁡2θ)(sec⁡2θ+tan⁡2θ)=1 (sec⁡2θ+tan⁡2θ).\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±trig1\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.