Ignoring principal value range Very Common CONCEPT
Students assume sin inverse(sin x) = x for all x, forgetting that the output must be in [-pi/2, pi/2].
Why: Students treat inverse trig functions as simple 'undo' operations without considering the restricted range.
WRONG: sin − 1 ( sin ( 5 π / 6 ) ) = 5 π / 6 \sin^{-1}(\sin(5\pi/6)) = 5\pi/6 sin − 1 ( sin ( 5 π /6 )) = 5 π /6 . This is wrong because 5 π / 6 ∉ [ − π / 2 , π / 2 ] 5\pi/6 \notin [-\pi/2, \pi/2] 5 π /6 ∈ / [ − π /2 , π /2 ] . RIGHT: sin ( 5 π / 6 ) = sin ( π / 6 ) = 1 / 2 \sin(5\pi/6) = \sin(\pi/6) = 1/2 sin ( 5 π /6 ) = sin ( π /6 ) = 1/2 . So sin − 1 ( sin ( 5 π / 6 ) ) = sin − 1 ( 1 / 2 ) = π / 6 \sin^{-1}(\sin(5\pi/6)) = \sin^{-1}(1/2) = \pi/6 sin − 1 ( sin ( 5 π /6 )) = sin − 1 ( 1/2 ) = π /6 . Always check that the answer lies in the principal range. See pattern: Find the Principal Value → Missing the xy condition in tan inverse sum Very Common FORMULA
Students use tan inverse(x) + tan inverse(y) = tan inverse((x+y)/(1-xy)) without checking whether xy < 1.
Why: The base formula is memorable, but the branch conditions (add pi when xy > 1) are often forgotten.
WRONG: tan − 1 2 + tan − 1 3 = tan − 1 2 + 3 1 − 6 = tan − 1 ( − 1 ) = − π / 4 \tan^{-1}2 + \tan^{-1}3 = \tan^{-1}\frac{2+3}{1-6} = \tan^{-1}(-1) = -\pi/4 tan − 1 2 + tan − 1 3 = tan − 1 1 − 6 2 + 3 = tan − 1 ( − 1 ) = − π /4 . This gives a negative answer for a sum of two positive values. RIGHT: Since x y = 6 > 1 xy = 6 > 1 x y = 6 > 1 and both x , y > 0 x, y > 0 x , y > 0 : tan − 1 2 + tan − 1 3 = π + tan − 1 ( − 1 ) = π − π / 4 = 3 π / 4 \tan^{-1}2 + \tan^{-1}3 = \pi + \tan^{-1}(-1) = \pi - \pi/4 = 3\pi/4 tan − 1 2 + tan − 1 3 = π + tan − 1 ( − 1 ) = π − π /4 = 3 π /4 . See pattern: Sum and Difference of Two Inverse Trig Values → Wrong domain in 2 tan inverse x identities Common FORMULA
Students apply 2*tan inverse(x) = sin inverse(2x/(1+x^2)) for all x, but this is valid only when |x| <= 1.
Why: The substitution x = tan(theta) gives theta in (-pi/2, pi/2), but 2*theta must also stay in the principal range of sin inverse.
WRONG: Applying 2 tan − 1 2 = sin − 1 ( 4 / 5 ) 2\tan^{-1}2 = \sin^{-1}(4/5) 2 tan − 1 2 = sin − 1 ( 4/5 ) directly. Since ∣ x ∣ = 2 > 1 |x| = 2 > 1 ∣ x ∣ = 2 > 1 , this identity does not hold. RIGHT: For ∣ x ∣ > 1 |x| > 1 ∣ x ∣ > 1 : 2 tan − 1 x = π − sin − 1 2 x 1 + x 2 2\tan^{-1}x = \pi - \sin^{-1}\frac{2x}{1+x^2} 2 tan − 1 x = π − sin − 1 1 + x 2 2 x (when x > 1 x > 1 x > 1 ). Use the cos − 1 \cos^{-1} cos − 1 form instead: 2 tan − 1 x = cos − 1 1 − x 2 1 + x 2 2\tan^{-1}x = \cos^{-1}\frac{1-x^2}{1+x^2} 2 tan − 1 x = cos − 1 1 + x 2 1 − x 2 for x ≥ 0 x \ge 0 x ≥ 0 . See pattern: Simplify Using Substitution → Not verifying solutions in inverse trig equations Common FORMULA
After solving an inverse trig equation algebraically, students skip checking whether the solutions satisfy the original domain constraints.
Why: Applying tan to both sides or squaring introduces extraneous solutions. The domain of each inverse trig term must be verified.
WRONG: Solving tan − 1 x + tan − 1 ( x + 1 ) = π / 4 \tan^{-1}x + \tan^{-1}(x+1) = \pi/4 tan − 1 x + tan − 1 ( x + 1 ) = π /4 and accepting both roots of the resulting quadratic without checking domains. RIGHT: After finding the quadratic roots, substitute each back into the original equation. Check that each argument lies in the valid domain and the equation holds with correct branch selection.
See pattern: Solve Inverse Trig Equations → Forgetting to split cases in sin inverse(2x sqrt(1-x^2)) Common FORMULA
Students apply sin inverse(2x*sqrt(1-x^2)) = 2*sin inverse(x) for all x in [-1,1], missing the case split at x = 1/sqrt(2).
Why: The formula looks clean without the case split, and students do not realize that 2*theta exceeds pi/2 when theta > pi/4.
WRONG: Writing sin − 1 ( 2 ⋅ 3 2 ⋅ 1 2 ) = 2 sin − 1 3 2 = 2 ⋅ π 3 = 2 π 3 \sin^{-1}(2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}) = 2\sin^{-1}\frac{\sqrt{3}}{2} = 2 \cdot \frac{\pi}{3} = \frac{2\pi}{3} sin − 1 ( 2 ⋅ 2 3 ⋅ 2 1 ) = 2 sin − 1 2 3 = 2 ⋅ 3 π = 3 2 π . But 2 π / 3 > π / 2 2\pi/3 > \pi/2 2 π /3 > π /2 , so this cannot be a principal value of sin inverse. RIGHT: Since x = 3 / 2 > 1 / 2 x = \sqrt{3}/2 > 1/\sqrt{2} x = 3 /2 > 1/ 2 : sin − 1 ( 2 x 1 − x 2 ) = π − 2 sin − 1 x = π − 2 π / 3 = π / 3 \sin^{-1}(2x\sqrt{1-x^2}) = \pi - 2\sin^{-1}x = \pi - 2\pi/3 = \pi/3 sin − 1 ( 2 x 1 − x 2 ) = π − 2 sin − 1 x = π − 2 π /3 = π /3 . See pattern: Simplify Using Substitution → Confusing inverse trig with reciprocal trig Occasional CONCEPT
Students write sin inverse(x) = 1/sin(x), confusing the inverse function with the reciprocal.
Why: The notation sin^(-1)(x) looks like sin(x) raised to the power -1, which suggests 1/sin(x) = cosec(x).
WRONG: sin − 1 ( 1 / 2 ) = 1 / sin ( 1 / 2 ) = cosec ( 1 / 2 ) \sin^{-1}(1/2) = 1/\sin(1/2) = \cosec(1/2) sin − 1 ( 1/2 ) = 1/ sin ( 1/2 ) = cosec ( 1/2 ) . RIGHT: sin − 1 ( 1 / 2 ) \sin^{-1}(1/2) sin − 1 ( 1/2 ) means 'the angle whose sine is 1 / 2 1/2 1/2 ', which is π / 6 \pi/6 π /6 . The reciprocal of sine is cosec, not sin inverse. See pattern: Find the Principal Value →