Does $\cos(x+y)=\cos x + \cos y$?
Find the value using a calculator: $\cos 75°$
At first I thought all I need is to separate the simpler known values like this:
$\cos 75^\circ = \cos 30°+\cos45° = {\sqrt3}/{2} + {\sqrt2}/{2} $
$= {(\sqrt3+\sqrt2)}/{2} $ This is my answer which translates to= $1.5731$ by calculator
However, when I used the calculator directly on $\cos 75°$, I get $0.2588$.
Where I am going wrong?
$\endgroup$ 67 Answers
$\begingroup$You used a formula $\cos(x+y) = \cos(x) + \cos(y)$ which is false. The correct formula is: $$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$
$\endgroup$ $\begingroup$The answer is simply: No and to see this take $y=0$ we find $$\cos(x)=\cos(x)+1$$ which's obviously false.
$\endgroup$ 0 $\begingroup$You can simply plot $cos(x+y)-(cos(x)+cos(y))$ to have your answer:
Alternatively: cos(A+B) = cosAcosB-sinAsinB with A = 45 and B is 30
$\endgroup$ 1 $\begingroup$As mentioned in other answers, $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$.
We also have $\cos(x)+\cos(y)=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$.
Since $\cos(0)=1$, we get $\cos(0+0)=1\ne2=\cos(0)+\cos(0)$.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$If you consider all $x,y$ in $\mathbb R$ such that $\cos(x+y)=\cos(x)+\cos(y)$ holds true, you get this.
