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Ptolemy's theorem states that the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. When those side-lengths are expressed in terms of the sin and cos values shown in the figure above, this yields the angle sum trigonometric identity for sine: sin(α + β) = sin α cos β + cos α sin β.
For the sine function, we can handle other values. If θ > π /2, then θ > 1. But sin θ ≤ 1 (because of the Pythagorean identity), so sin θ < θ. So we have < <. For negative values of θ we have, by the symmetry of the sine function
[1] [10] Another precarious convention used by a small number of authors is to use an uppercase first letter, along with a “ −1 ” superscript: Sin −1 (x), Cos −1 (x), Tan −1 (x), etc. [11] Although it is intended to avoid confusion with the reciprocal, which should be represented by sin −1 (x), cos −1 (x), etc., or, better, by ...
The real part of the other side is a polynomial in cos x and sin x, in which all powers of sin x are even and thus replaceable through the identity cos 2 x + sin 2 x = 1. By the same reasoning, sin nx is the imaginary part of the polynomial, in which all powers of sin x are odd and thus, if one factor of sin x is factored out, the remaining ...
2.3 Trigonometric, inverse trigonometric, hyperbolic, and inverse hyperbolic functions relationship 2.4 Modified-factorial denominators 2.5 Binomial coefficients
[1] [2] One reason for this is that they can greatly simplify differential equations that do not need to be answered with absolute precision. There are a number of ways to demonstrate the validity of the small-angle approximations. The most direct method is to truncate the Maclaurin series for each of the trigonometric functions.
Alternatively, the identities found at Trigonometric symmetry, shifts, and periodicity may be employed. By the periodicity identities we can say if the formula is true for −π < θ ≤ π then it is true for all real θ. Next we prove the identity in the range π / 2 < θ ≤ π.
Trigonometric identities may help simplify the answer. [1] [2] Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the antiderivative before applying the boundaries of integration.