How does $(1+\sin x)\cos x =\cos x+\sin x \cos x$? I have a feeling i missed a basic fact Okay, so i have problem: $\cos x+ \sin (2x) = 0$. When I searched for help on the internet, everyone seemed to use that $(1+\sin x)\cos x =\cos x+\sin x \cos x$. Is this a formula that I have missed requiring a hard proof? Or am I simply missing some basic algebra?


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*Mads

 A: This is the distributive property: $(A + B) \times C = A \times C + B \times C$.  In your case, $A$ is $1$, $B$ is $\sin(x)$, and $C$ is $\cos(x)$.
A: Notice, this is not any formula. It's just multiplication by distribution. One should multiply by distributing $\cos x$ over $(1+\sin x)$ as follows $$(1+\sin x)\cos x=\cos x+\sin x\cdot \cos x$$
Or in general one should know multiplication by distribution  $$(X+Y)Z=X\cdot Z+Y\cdot Z$$
A: For every $x \in \mathbb R$, $sin x$ and $cosx$ are real numbers. In the real numbers the operation of multiplication is distributive, so you can use the rule
(a+b)c=ac + bc = c(a+b).
Even more the multiplication is commutative so $ac=ca$ and also addiction is commutative so $a+b= b+a$. At the very end, you can rewrite your formula in so many different ways:
$(1+\sin x)\cos x = \cos x+\sin x \cos x = $
$=\cos x(1+\sin x) = \cos x(\sin x + 1)=$
$= \cos x \sin x + \cos x = \sin x \cos x + \cos x$
and so on. All of them are true of course for every $x \in \mathbb R$.
A: It is a term by term multiplication in the title.
Your problem and solution
$$ \sin x + \sin 2 x =0 ; \sin x ( 1 + 2 \cos x) = 0;  x=0, \pi... 2 \pi/3, 5 \pi/3 ... $$ 
