Why an inverse needs a restricted domain. A function has an inverse only where it is one-to-one and onto. None of the six trigonometric functions is one-to-one on its natural domain, because each is periodic — infinitely many angles give the same sine, the same cosine, and so on. (Compare y=x2 on all of R: it isn't one-to-one either, since x=2 and x=−2 both give y=4; only once the domain is cut down to x≥0 does an inverse, x, exist.) So before "sin−1", "cos−1", etc. can mean anything, each trigonometric function is first restricted to one conventional interval on which it is one-to-one and whose image is its whole natural range. That restricted interval always contains 0 together with some positive angles.
The six function/inverse domain-range pairs.
Function (restricted)
Domain → Range
Inverse
Domain → Range
sinx
[−2π,2π]→[−1,1]
sin−1x
[−1,1]→[−2π,2π]
cosx
[0,π]→[−1,1]
cos−1x
[−1,1]→[0,π]
tanx
(−2π,2π)→(−∞,∞)
tan−1x
(−∞,∞)→(−2π,2π)
cotx
(0,π)→(−∞,∞)
cot−1x
(−∞,∞)→(0,π)
cscx
[−2π,2π]−{0}→R−(−1,1)
csc−1x
R−(−1,1)→[−2π,2π]−{0}
secx
[0,π]−{2π}→R−(−1,1)
sec−1x
R−(−1,1)→[0,π]−{2π}
Each inverse function's stated domain is exactly the range of the original restricted function, and its range is exactly that restricted domain — an inverse always swaps domain and range with the function it undoes, and its graph is the mirror image of the original's graph in the line y=x.
The principal value. For y=sinx, a given value t∈[−1,1] is achieved by infinitely many angles, but only one of them lies in [−2π,2π]. That one angle is called the principal value, written sin−1t, and the same idea applies to each of the other five inverse functions using its own restricted interval. Framed generally: among every angle that satisfies the equation, the principal value is the one numerically smallest in magnitude — it may be positive or negative. If the smallest-magnitude candidates come in a +θ,−θ pair (numerically equal, opposite sign), the convention is to choose the positive one. The sign of the input further pins down which half of the inverse function's interval the principal value sits in (e.g. sin−1x≥0 when x≥0, and sin−1x<0 when x<0) — this pattern, tabulated for all six functions, is what makes evaluating a principal value a matter of pattern-matching to a known angle rather than guessing.
Two clarifications worth keeping distinct: sin−1x is notsinx1 — the −1 is purely an inverse-function label; and sin−1x is also written arcsinx (similarly arccos,arctan,…).
Worked illustration. Find the principal value of cos−1(−21). Let y=cos−1(−21), so by definition y∈[0,π] and cosy=−21. Since cos3π=21, and cosine is negative in the second quadrant with cos(π−θ)=−cosθ, we get cos(π−3π)=−21, i.e. cos32π=−21. Because 32π lies in [0,π], it is the required principal value: cos−1(−21)=32π. (This also matches the sign rule above: the input is negative, so the principal value of cos−1 must land in (2π,π] — and 32π does.)
Set each expression equal to y in the inverse function's own restricted interval, and match the value to a known angle in that interval.
✓Final answer
(i) 4π (ii) 6π (iii) −2π (iv) 43π (v) 3π.
Each part is solved the same way: write y= (the inverse expression) with y restricted to that function's principal-value interval, convert to the direct trig equation, and identify the matching standard angle inside the interval.
Step 1. Part (i) sin−121. Let y=sin−121, y∈[−2π,2π]. Then siny=21=sin4π, and 4π lies in the interval, so y=4π.
Step 2. Part (ii) cos−123. Let y=cos−123, y∈[0,π]. Then cosy=23=cos6π, and 6π∈[0,π], so y=6π.
Step 3. Part (iii) csc−1(−1). Let y=csc−1(−1), y∈[−2π,2π]−{0}. Then cscy=−1⇒siny=−1, which happens at y=−2π, and this lies in the interval, so y=−2π.
Step 4. Part (iv) sec−1(−2). Let y=sec−1(−2), y∈[0,π]−{2π}. Then secy=−2⇒cosy=−21. Since cos43π=−21 and 43π∈[0,π], y=43π.
Step 5. Part (v) tan−1(3). Let y=tan−1(3), y∈(−2π,2π). Then tany=3=tan3π, and 3π lies in the interval, so y=3π.
✓Final answer
(i) 4π (ii) 6π (iii) −2π (iv) 43π (v) 3π.
Set the expression =y in the function's principal-value interval and match a known angle
Picking a coterminal angle outside the function's own restricted interval (e.g. giving csc−1(−1)=23π instead of −2π)
Forgetting sec−1 and csc−1 convert to cos/sin via the reciprocal before matching the angle