$cos(a+b)=cos a cos b -sin a sin b$
$cos(2a)=cos^2a-sin^2a$
$cos(ab)=?$
$egingroup$ Are $a,b$ arbitrary, or are you assuming that $binriz15.combbZ$ is an integer? $endgroup$
No, và there"s a precise reason.
Bạn đang xem: Trigonometry: find the exact value of cos ( a
First, the geometric definition of $cos$ talks about angles, & the product of two angles doesn"t make sense.
Moreover, when you view the cosine as an exponential complex function, as you know $$cosx= frac e^i x + e^-i x2 $$ you can see that the identities you quoted come from properties of powers, such as $e^a+b=e^a e^b$ or $e^2a = (e^a)^2$
Since there"s no significant formula for $e^ab$, there isn"t one for the $cos$ function too.
If $a$ is an integer and $b$ is an angle,
$$cos(ab) = T_a(cos b)$$
where $T_n(x)$ is the $n^th$ Chebyshev polynomial.
Not really, but I suppose this works:$$cos ab=Re<(cos(b)+isin(b))^a>$$
You can get the above equation by taking the real part of de Moivre"s formula:$$cos n heta +isin n heta=(cos( heta)+isin( heta))^n ,$$
For general $a$ & $b$, we cannot write $cos (ab)$ in terms of the trig functions $cos a,sin a, cos b, sin b$. This is because the trig functions are periodic with period $2pi$, so adding $2pi$ to lớn $b$ does not change any of these functions. But adding $2pi$ to $b$ can change $cos (ab)$ - for instance, if $a=1/2$, if sends $cos (ab)$ khổng lồ $-cos(ab)$. Only if $a$ is an integer can we avoid this problem.
As many experts already noted here, an argument of cos(⋯) is an angle, and a sensible briz15.comematical structure on angles is the one of abelian-groups under “+”. We can showroom and subtract angles, as well as multiply them by integers. To lớn some extent we can multiply angles by rational numbers, i.e. Solve equations like$$ qx = pa,quad x,a ext are angles, p,qinriz15.combb Z, q≠0.$$If $a$ is specified modulo 2π radians, then such solutions, placed on the trigonometric circle, will khung a set of $q$ elements (vertices of a regular $q$-gon), that can be described with an algebraic equation.
There is no “reasonable” multiplication of an angle & an irrational number $t$. If $a$ is specified modulo 2π radians (that is a typical condition), then possible values of $ta$ will khung a dense subset of the trigonometric circle, & hence values of trigonometric functions on it will have no use for calculations.
Conclusion
There are trigonometric identities for products of an angle and a rational number; see other answers và this page for some partial cases.
Xem thêm: Tìm Từ Đồng Nghĩa Chỉ Màu Xanh, Tìm Các Từ Đồng Nghĩa Chỉ Các Màu Sắc Dưới Đây
There are no products of an angle và an irrational number, as well as there are no products of two angles.