Principal values Since none of the six trigonometric functions are
one-to-one, they must be restricted in order to have inverse functions. Therefore, the result
ranges of the inverse functions are proper (i.e. strict)
subsets of the domains of the original functions. For example, using in the sense of
multivalued functions, just as the
square root function y = \sqrt{x} could be defined from y^2 = x, the function y = \arcsin(x) is defined so that \sin(y) = x. For a given
real number x, with -1 \leq x \leq 1, there are multiple (in fact,
countably infinitely many) numbers y such that \sin(y) = x; for example, \sin(0) = 0, but also \sin(\pi) = 0, \sin(2 \pi) = 0, etc. When only one value is desired, the function may be restricted to its
principal branch. With this restriction, for each x in the domain, the expression \arcsin(x) will evaluate only to a single value, called its
principal value. These properties apply to all the inverse trigonometric functions. The principal inverses are listed in the following table. Note: Some authors define the range of arcsecant to be {{nowrap|(0 \leq y }} or {{nowrap|\pi \leq y ),}} because the tangent function is nonnegative on this domain. This makes some computations more consistent. For example, using this range, \tan(\arcsec(x)) = \sqrt{x^2 - 1}, whereas with the range {{nowrap|(0 \leq y }} or {{nowrap|\frac{\pi}{2} ),}} we would have to write \tan(\arcsec(x)) = \pm \sqrt{x^2 - 1}, since tangent is nonnegative on 0 \leq y but nonpositive on \frac{\pi}{2} For a similar reason, the same authors define the range of arccosecant to be ( - \pi or 0
Domains If is allowed to be a
complex number, then the range of applies only to its real part.
Solutions to elementary trigonometric equations Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of 2 \pi: • Sine and cosecant begin their period at 2 \pi k-\frac{\pi}{2} (where k is an integer), finish it at 2 \pi k+\frac{\pi}{2}, and then reverse themselves over 2 \pi k+\frac{\pi}{2} to 2 \pi k+\frac{3\pi}{2}. • Cosine and secant begin their period at 2 \pi k, finish it at 2 \pi k+\pi. and then reverse themselves over 2 \pi k+\pi to 2 \pi k+2 \pi. • Tangent begins its period at 2 \pi k-\frac{\pi}{2}, finishes it at 2 \pi k+\frac{\pi}{2}, and then repeats it (forward) over 2 \pi k+\frac{\pi}{2} to 2 \pi k+\frac{3 \pi}{2}. • Cotangent begins its period at 2 \pi k, finishes it at 2 \pi k+\pi, and then repeats it (forward) over 2 \pi k+\pi to 2 \pi k+2 \pi. This periodicity is reflected in the general inverses, where k is some integer. The following table shows how inverse trigonometric functions may be used to solve equalities involving the six standard trigonometric functions. It is assumed that the given values \theta, r, s, x, and y all lie within appropriate ranges so that the relevant expressions below are
well-defined. Note that "for some k \in \Z" is just another way of saying "for some
integer k." The symbol \,\iff\, is
logical equality and indicates that if the left hand side is true then so is the right hand side and, conversely, if the right hand side is true then so is the left hand side (see this footnote for more details and an example illustrating this concept). where the first four solutions can be written in expanded form as: For example, if \cos \theta = -1 then \theta = \pi+2 \pi k = -\pi+2 \pi (1+k) for some k \in \Z. While if \sin \theta = \pm 1 then \theta = \frac{\pi}{2}+\pi k =-\frac{\pi}{2}+\pi (k+1) for some k \in \Z, where k will be even if \sin \theta = 1 and it will be odd if \sin \theta = -1. The equations \sec \theta = -1 and \csc \theta = \pm 1 have the same solutions as \cos \theta = -1 and \sin \theta = \pm 1, respectively. In all equations above for those just solved (i.e. except for \sin/\csc \theta = \pm 1 and \cos/\sec \theta =-1), the integer k in the solution's formula is uniquely determined by \theta (for fixed r, s, x, and y). With the help of
integer parity \operatorname{Parity}(h) = \begin{cases} 0 & \text{if } h \text{ is even } \\ 1 & \text{if } h \text{ is odd } \\ \end{cases} it is possible to write a solution to \cos \theta = x that doesn't involve the "plus or minus" \,\pm\, symbol: :cos \; \theta = x \quad if and only if \quad \theta = (-1)^h \arccos(x) + \pi h + \pi \operatorname{Parity}(h) \quad for some h \in \Z. And similarly for the secant function, :sec \; \theta = r \quad if and only if \quad \theta = (-1)^h \arcsec(r) + \pi h + \pi \operatorname{Parity}(h) \quad for some h \in \Z, where \pi h + \pi \operatorname{Parity}(h) equals \pi h when the integer h is even, and equals \pi h + \pi when it's odd.
Detailed example and explanation of the "plus or minus" symbol The solutions to \cos \theta = x and \sec \theta = x involve the "plus or minus" symbol \,\pm,\, whose meaning is now clarified. Only the solution to \cos \theta = x will be discussed since the discussion for \sec \theta = x is the same. We are given x between -1 \leq x \leq 1 and we know that there is an angle \theta in some interval that satisfies \cos \theta = x. We want to find this \theta. The table above indicates that the solution is \,\theta = \pm \arccos x+2 \pi k\, \quad \text{ for some }k \in \Z which is a shorthand way of saying that (at least) one of the following statement is true: • \,\theta = \arccos x+2 \pi k\, for some integer k, or • \,\theta =-\arccos x+2 \pi k\, for some integer k. As mentioned above, if \,\arccos x = \pi\, (which by definition only happens when x = \cos \pi = -1) then both statements (1) and (2) hold, although with different values for the integer k: if K is the integer from statement (1), meaning that \theta = \pi+2 \pi K holds, then the integer k for statement (2) is K+1 (because \theta = -\pi+2 \pi (1+K)). However, if x \neq -1 then the integer k is unique and completely determined by \theta. If \,\arccos x = 0\, (which by definition only happens when x = \cos 0 = 1) then \,\pm\arccos x = 0\, (because \,+ \arccos x = +0 = 0\, and \,-\arccos x = -0 = 0\, so in both cases \,\pm \arccos x\, is equal to 0) and so the statements (1) and (2) happen to be identical in this particular case (and so both hold). Having considered the cases \,\arccos x = 0\, and \,\arccos x = \pi,\, we now focus on the case where \,\arccos x \neq 0\, and \,\arccos x \neq \pi,\, So assume this from now on. The solution to \cos \theta = x is still \,\theta = \pm \arccos x+2 \pi k\, \quad \text{ for some }k \in \Z which as before is shorthand for saying that one of statements (1) and (2) is true. However this time, because \,\arccos x \neq 0\, and \,0 statements (1) and (2) are different and furthermore,
exactly one of the two equalities holds (not both). Additional information about \theta is needed to determine which one holds. For example, suppose that x = 0 and that that is known about \theta is that \,-\pi \leq \theta \leq \pi\, (and nothing more is known). Then \arccos x = \arccos 0 = \frac{\pi}{2} and moreover, in this particular case k = 0 (for both the \,+\, case and the \,-\, case) and so consequently, \theta ~=~ \pm \arccos x+2 \pi k ~=~ \pm \left(\frac{\pi}{2}\right)+2\pi (0) ~=~ \pm \frac{\pi}{2}. This means that \theta could be either \,\pi/2\, or \,-\pi/2. Without additional information it is not possible to determine which of these values \theta has. An example of some additional information that could determine the value of \theta would be knowing that the angle is above the x-axis (in which case \theta = \pi/2) or alternatively, knowing that it is below the x-axis (in which case \theta =-\pi/2).
Equal identical trigonometric functions ;Set of all solutions to elementary trigonometric equations Thus given a single solution \theta to an elementary trigonometric equation (\sin \theta = y is such an equation, for instance, and because \sin (\arcsin y) = y always holds, \theta := \arcsin y is always a solution), the set of all solutions to it are:
Transforming equations The equations above can be transformed by using the reflection and shift identities: These formulas imply, in particular, that the following hold: \begin{align} \sin \theta &= -\sin(-\theta) &&= -\sin(\pi+\theta) &&= \phantom{-}\sin(\pi-\theta) \\ &= -\cos\left(\frac{\pi}{2}+\theta\right) &&= \phantom{-}\cos\left(\frac{\pi}{2}-\theta\right) &&= -\cos\left(-\frac{\pi}{2}-\theta\right) \\ &= \phantom{-}\cos\left(-\frac{\pi}{2}+\theta\right) &&= -\cos\left(\frac{3\pi}{2}-\theta\right) &&= -\cos\left(-\frac{3\pi}{2}+\theta\right) \\[0.3ex] \cos \theta &= \phantom{-}\cos(-\theta) &&= -\cos(\pi+\theta) &&= -\cos(\pi-\theta) \\ &= \phantom{-}\sin\left(\frac{\pi}{2}+\theta\right) &&= \phantom{-}\sin\left(\frac{\pi}{2}-\theta\right) &&= -\sin\left(-\frac{\pi}{2}-\theta\right) \\ &= -\sin\left(-\frac{\pi}{2}+\theta\right) &&= -\sin\left(\frac{3\pi}{2}-\theta\right) &&= \phantom{-}\sin\left(-\frac{3\pi}{2}+\theta\right) \\[0.3ex] \tan \theta &= -\tan(-\theta) &&= \phantom{-}\tan(\pi+\theta) &&= -\tan(\pi-\theta) \\ &= -\cot\left(\frac{\pi}{2}+\theta\right) &&= \phantom{-}\cot\left(\frac{\pi}{2}-\theta\right) &&= \phantom{-}\cot\left(-\frac{\pi}{2}-\theta\right) \\ &= -\cot\left(-\frac{\pi}{2}+\theta\right) &&= \phantom{-}\cot\left(\frac{3\pi}{2}-\theta\right) &&= -\cot\left(-\frac{3\pi}{2}+\theta\right) \\[0.3ex] \end{align} where swapping \sin \leftrightarrow \csc, swapping \cos \leftrightarrow \sec, and swapping \tan \leftrightarrow \cot gives the analogous equations for \csc, \sec, \text{ and } \cot, respectively. So for example, by using the equality \sin \left(\frac{\pi}{2}-\theta\right) = \cos \theta, the equation \cos \theta = x can be transformed into \sin \left(\frac{\pi}{2}-\theta\right) = x, which allows for the solution to the equation \;\sin \varphi = x\; (where \varphi := \frac{\pi}{2}-\theta) to be used; that solution being: \varphi = (-1)^k \arcsin (x)+\pi k \; \text{ for some } k \in \Z, which becomes: \frac{\pi}{2}-\theta ~=~ (-1)^k \arcsin (x)+\pi k \quad \text{ for some } k \in \Z where using the fact that (-1)^{k} = (-1)^{-k} and substituting h :=-k proves that another solution to \;\cos \theta = x\; is: \theta ~=~ (-1)^{h+1} \arcsin (x)+\pi h+\frac{\pi}{2} \quad \text{ for some } h \in \Z. The substitution \;\arcsin x = \frac{\pi}{2}-\arccos x\; may be used express the right hand side of the above formula in terms of \;\arccos x\; instead of \;\arcsin x.\;
Relationships between trigonometric functions and inverse trigonometric functions Trigonometric functions of inverse trigonometric functions are tabulated below. They may be derived from the Pythagorean identities. Another way is by considering the geometry of a right-angled triangle, with one side of length 1 and another side of length x, then applying the
Pythagorean theorem and definitions of the trigonometric ratios. It is worth noting that for arcsecant and arccosecant, the diagram assumes that x is positive, and thus the result has to be corrected through the use of
absolute values and the
signum (sgn) operation.
Relationships among the inverse trigonometric functions Complementary angles: :\begin{align} \arccos(x) &= \frac{\pi}{2} - \arcsin(x) \\[0.5em] \arccot(x) &= \frac{\pi}{2} - \arctan(x) \\[0.5em] \arccsc(x) &= \frac{\pi}{2} - \arcsec(x) \end{align} Negative arguments: :\begin{align} \arcsin(-x) &= -\arcsin(x) \\ \arccsc(-x) &= -\arccsc(x) \\ \arccos(-x) &= \pi -\arccos(x) \\ \arcsec(-x) &= \pi -\arcsec(x) \\ \arctan(-x) &= -\arctan(x) \\ \arccot(-x) &= \pi -\arccot(x) \end{align} Reciprocal arguments: :\begin{align} \arcsin\left(\frac{1}{x}\right) &= \arccsc(x) & \\[0.3em] \arccsc\left(\frac{1}{x}\right) &= \arcsin(x) & \\[0.3em] \arccos\left(\frac{1}{x}\right) &= \arcsec(x) & \\[0.3em] \arcsec\left(\frac{1}{x}\right) &= \arccos(x) & \\[0.3em] \arctan\left(\frac{1}{x}\right) &= \arccot(x) &= \frac{\pi}{2} - \arctan(x) \, , \text{ if } x > 0 \\[0.3em] \arctan\left(\frac{1}{x}\right) &= \arccot(x) - \pi &= -\frac{\pi}{2} - \arctan(x) \, , \text{ if } x 0 \\[0.3em] \arccot\left(\frac{1}{x}\right) &= \arctan(x) + \pi &= \frac{3\pi}{2} - \arccot(x) \, , \text{ if } x The identities above can be used with (and derived from) the fact that \sin and \csc are
reciprocals (i.e. \csc = \tfrac1{\sin}), as are \cos and \sec, and \tan and \cot. Useful identities if one only has a fragment of a sine table: :\begin{align} \arcsin(x) &= \frac{1}{2}\arccos\left(1-2x^2\right) \, , \text{ if } 0 \leq x \leq 1 \\ \arcsin(x) &= \arctan\left(\frac{x}{\sqrt{1 - x^2}}\right) \\ \arccos(x) &= \frac{1}{2}\arccos\left(2x^2-1\right) \, , \text{ if } 0 \leq x \leq 1 \\ \arccos(x) &= \arctan\left(\frac{\sqrt{1 - x^2}}{x}\right) \\ \arccos(x) &= \arcsin\left(\sqrt{1 - x^2}\right) \, , \text{ if } 0 \leq x \leq 1 \text{ , from which you get } \\ \arccos &\left(\frac{1-x^2}{1 + x^2}\right) = \arcsin \left (\frac{2x}{1 + x^2}\right) \, , \text{ if } 0 \leq x \leq 1 \\ \arcsin &\left(\sqrt{1 - x^2}\right) =\frac{\pi}{2}-\sgn(x)\arcsin(x) \\ \arctan(x) &= \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right) \\ \arccot(x) &= \arccos\left(\frac{x}{\sqrt{1 + x^2}}\right) \end{align} Whenever the square root of a complex number is used here, we choose the root with the positive real part (or positive imaginary part if the square was negative real). A useful form that follows directly from the table above is :\arctan(x) = \arccos\left(\sqrt{\frac{1}{1+x^2}}\right)\, , \text{ if } x\geq 0 . It is obtained by recognizing that \cos\left(\arctan\left(x\right)\right) = \sqrt{\frac{1}{1+x^2}} = \cos\left(\arccos\left(\sqrt{\frac{1}{1+x^2}}\right)\right). From the
half-angle formula, \tan\left(\tfrac{\theta}{2}\right) = \tfrac{\sin(\theta)}{1 + \cos(\theta)}, we get: :\begin{align} \arcsin(x) &= 2 \arctan\left(\frac{x}{1 + \sqrt{1 - x^2}}\right) \\[0.5em] \arccos(x) &= 2 \arctan\left(\frac{\sqrt{1 - x^2}}{1 + x}\right) \, , \text{ if } -1
Arctangent addition formula :\arctan(u) \pm \arctan(v) = \arctan\left(\frac{u \pm v}{1 \mp uv}\right) \pmod \pi \, , \quad u v \ne 1 \, . This is derived from the tangent
addition formula :\tan(\alpha \pm \beta) = \frac{\tan(\alpha) \pm \tan(\beta)}{1 \mp \tan(\alpha)\tan(\beta)} \, , by letting :\alpha = \arctan(u) \, , \quad \beta = \arctan(v) \, . ==In calculus==