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This geometric argument relies on definitions of arc length and area, which act as assumptions, so it is rather a condition imposed in construction of trigonometric functions than a provable property. [2] 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 ...
The fact that the triple-angle formula for sine and cosine only involves powers of a single function allows one to relate the geometric problem of a compass and straightedge construction of angle trisection to the algebraic problem of solving a cubic equation, which allows one to prove that trisection is in general impossible using the given tools.
The tangent of half an angle is important in spherical trigonometry and was sometimes known in the 17th century as the half tangent or semi-tangent. [2] Leonhard Euler used it to evaluate the integral ∫ d x / ( a + b cos x ) {\textstyle \int dx/(a+b\cos x)} in his 1768 integral calculus textbook , [ 3 ] and Adrien-Marie Legendre described ...
[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.
The angle between the horizontal line and the shown diagonal is 1 / 2 (a + b). This is a geometric way to prove the particular tangent half-angle formula that says tan 1 / 2 (a + b) = (sin a + sin b) / (cos a + cos b). The formulae sin 1 / 2 (a + b) and cos 1 / 2 (a + b) are the ratios of the actual distances to ...
To prove the law of tangents one can start with the law of sines: a sin α = b sin β = d , {\displaystyle {\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}=d,} where d {\displaystyle d} is the diameter of the circumcircle , so that a = d sin α {\displaystyle a=d\sin \alpha } and b = d sin β {\displaystyle ...
satisfying respectively y(0) = 0, y ′ (0) = 1 and y(0) = 1, y ′ (0) = 0. It follows from the theory of ordinary differential equations that the first solution, sine, has the second, cosine, as its derivative, and it follows from this that the derivative of cosine is the negative of the sine. The identity is equivalent to the assertion that ...
We conclude that for 0 < θ < 1 / 2 π, the quantity sin(θ)/θ is always less than 1 and always greater than cos(θ). Thus, as θ gets closer to 0, sin(θ)/θ is "squeezed" between a ceiling at height 1 and a floor at height cos θ, which rises towards 1; hence sin(θ)/θ must tend to 1 as θ tends to 0 from the positive side: