2016年4月27日 · Opposite real number identities(why $\cos(-x)=\cos x$) 6 Given three chords on a circle which form an interior triangle what is the area of the triangle formed?

2017年4月17日 · The equation in question is a transcendental equation.Apart of guessing, numerical or analytical methods, there is no way of solving the equation without using another transcendental function, and therefore argue in circles.

2012年4月13日 · There was a proof that $\cos^{(3)}\sinh x=\sin^{(3)}\cosh x$ has infinitely many solutions in a previous version of this answer, but it turns out this is irrelevant to the question.

The maximum value of cos(x+y)=1. The maximum values of cos(x) and cos(y) are 1. Since cos(x)+cos(y)=2cos(x+y) for some values of x and y, it cannot be the case that cos(x)+cos(y)=cos(x+y) except for special values of x and y.

2018年4月13日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

In quadrant IV, $\cos{x}$ is positive, so taking a square root gives us $$\cos{x} = \sqrt{1 - \sin^2{x}}$$

2016年2月23日 · You don't (in terms of elementary functions). Unless a power series will do... $$\cos (1) (x-\pi ) +\frac{1}{6} \sin (1)(x-\pi )^3 + \left(-\frac{\sin (1)}{120}-\frac ...

2018年8月19日 · The sequence of $$ a_x ={\cos (x)\over x} $$ does converge to zero. As a result, intuitively $$ \sum_{x=1}^\infty {\cos (x)\over x} $$ should also converge right?

$\begingroup$ $\cos^2(x) = \cos(x)\times \cos(x)$ and $\cos(x^2) = \cos(x \times x)$ So no. But beware, the notation $\cos^{-1}(x)$ is ambiguous. It can denote the inverse cosine function or the reciprocal of the cosine function. $\endgroup$ –

2016年10月25日 · This is a proof We use the sum of angles formula: cos(A+-B) -= cosAcosB+-sinAsinB Put A=0 and B=-x to get cos(0-x) -= cos0cosx-sin0sinx And we know that sin0=0 and cos0=1 so: cos(-x) -= (1)cosx - 0 :. cos(-x) -= cosx QED