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Concept 11 / 11Trigonometry
Inverse Trigonometric Identities: Sums, Multiple Angles and Equations
The complementary identities sin^-1 x + cos^-1 x = pi/2, the addition formulas for tan^-1 with their range conditions, formulas for sin^-1 x +/- sin^-1 y, 2 tan^-1 x in terms of sin^-1, cos^-1 and tan^-1 with validity intervals, and solving equations and summing series with inverse trigonometric functions.
Complementary identities (for x in the relevant domains):
sin−1x+cos−1x=2π, tan−1x+cot−1x=2π, sec−1x+csc−1x=2π.
Reason: if θ=sin−1x, then cos(2π−θ)=x with 2π−θ∈[0,π], so 2π−θ=cos−1x.
Sum and difference of inverse tangents. From tan(A+B):
tan−1x+tan−1y=tan−11−xyx+y if xy<1;
=π+tan−11−xyx+y if xy>1 and x,y>0;
=−π+tan−11−xyx+y if xy>1 and x,y<0;
=2π if xy=1 with x,y>0.
tan−1x−tan−1y=tan−11+xyx−y if xy>−1 (with similar corrections otherwise).
The corrections are needed because the sum of two angles in (−2π,2π) can lie outside that interval, where tan−1 cannot reach. For example, tan−12+tan−13=π+tan−1−55=π−4π=43π, not −4π. Three terms: tan−1x+tan−1y+tan−1z=tan−11−xy−yz−zxx+y+z−xyz (with range adjustments).
Sums of inverse sines and cosines (for x,y≥0 with x2+y2≤1):
sin−1x+sin−1y=sin−1(x1−y2+y1−x2),
cos−1x+cos−1y=cos−1(xy−1−x21−y2).
When x2+y2>1 (with x,y>0), the sum of inverse sines exceeds 2π and the formula becomes π−sin−1(…).
Multiple-angle forms of 2tan−1x:
2tan−1x=sin−11+x22x for ∣x∣≤1;
2tan−1x=cos−11+x21−x2 for x≥0;
2tan−1x=tan−11−x22x for ∣x∣<1.
Also 2sin−1x=sin−1(2x1−x2) for ∣x∣≤21 and 3sin−1x=sin−1(3x−4x3) for ∣x∣≤21. Each is valid only where the left side stays inside the principal range of the right-side inverse. Outside those intervals the correct forms involve π± adjustments. These validity conditions are exactly what makes derivatives of such expressions change sign across ∣x∣=1 (see the chapter on differentiation).
Solving equations. Take a suitable trigonometric function of both sides (usually tan or cos), solve the algebraic equation, and check each root in the original equation, because taking tan of both sides can introduce angles differing by π.
Summing series. Write each term as a difference: tan−11+abb−a=tan−1b−tan−1a for a,b≥0. Terms like tan−11+n+n21 and tan−1n22 (=tan−11+(n+1)(n−1)(n+1)−(n−1)) telescope.
Key Takeaways:
- sin−1x+cos−1x=2π; tan−1x+cot−1x=2π.
- tan−1x+tan−1y=tan−11−xyx+y only when xy<1; add π when xy>1 with positive x,y.
- 2tan−1x=sin−11+x22x for ∣x∣≤1; =cos−11+x21−x2 for x≥0.
- 3sin−1x=sin−1(3x−4x3) for ∣x∣≤21.
- Check every root after taking a trigonometric function of both sides.
- Telescoping: tan−11+abb−a=tan−1b−tan−1a.
- Pitfall: applying the tan−1 addition formula without checking xy<1.
Brain-Teaser 1: Prove that tan−11+tan−12+tan−13=π.
Solution: tan−12+tan−13: here xy=6>1 with both positive, so it equals π+tan−11−65=π+tan−1(−1)=π−4π=43π. Adding tan−11=4π gives π. Using the formula without the π correction would give 0, which is impossible for a sum of three positive angles.
Brain-Teaser 2: Solve tan−1(x−1)+tan−1(x+1)=tan−13x.
Solution: Take tan of both sides: 1−(x2−1)2x=3x, i.e. 2−x22x=3x. So x=0 or 2=3(2−x2), giving x2=34, x=±32. Check x=32≈1.155: the left side is tan−10.155+tan−12.155≈0.154+1.136=1.290, and tan−13.464≈1.290. It works. For x=−32, by oddness both sides change sign, so it also works; x=0 works trivially. Solutions: 0,±32. (Here (x−1)(x+1)=31<1, so no range correction was needed, which the check confirms.)
Brain-Teaser 3: Find ∑n=1∞tan−1n22.
Solution: n22=1+(n+1)(n−1)(n+1)−(n−1), so each term is tan−1(n+1)−tan−1(n−1) (both arguments non-negative, so no correction). The partial sum to N telescopes, leaving tan−1(N+1)+tan−1N−tan−11−tan−10. As N→∞ this tends to 2π+2π−4π−0=43π. The gap of 2 in the telescoping means two terms survive at each end.
