Question regarding in periodic function

Question

I have question I know that $\cos(x+2\pi)=\cos x$ and $\sin(x+2\pi)=\sin x$ but if we have $\cos(x+\pi)=?$ and $\sin(x+\pi)=?$ with explaination thanks

Answer 1

It is not quite clear what kind of explanation you want, but using Euler's formula $$e^{ix}=\cos(x)+i\sin(x)$$ so $\cos(x+\pi)+i\sin(x+\pi)=e^{i(x+\pi)}=e^{ix}e^{i\pi}$ and $e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1$ so matching real and imaginary parts

$$\cos(x+\pi)=-\cos(x)$$ $$\sin(x+\pi)=-\sin(x)$$

Answer 2

There's a simple and efficient method to answer this kind of questions: draw the trigonometric circle and compare $\cos(x+\pi)$ with $\cos x, \sin x,\ldots$ and we can see that

$$\cos(x+\pi)=-\cos x\quad\text{and}\quad \sin(x+\pi)=-\sin x$$

Of course there's also an analytic proof.

Answer 3

Another approach, if you recall the the trig identities $$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$$ $$\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$$ then we can put $a=x$ and $b=\pi$ to obtain $$\cos(x+\pi) = \cos(x)\cos(\pi) - \sin(x)\sin(\pi) = \cos(x)(-1) - \sin(x)(0) = -\cos(x)$$ and $$\sin(x+\pi) = \sin(x)\cos(\pi) + \cos(x)\sin(\pi) = \sin(x)(-1) + \cos(x)(0) = -\sin(x)$$