Why is sin (x+x) = sinx cosx + cosx sinx ?

[Natasha1]

Why is sin (x+x) = sinx cosx + cosx sinx ? Simple explanation required please

 

[rindech]

sin (x+y) = sin x cos y + cos x sin y. Simply place an "x" for the "y" in the formula. Noted: sin (2x) = 2sin x cos x.

 

[LeonhardEuler]

I don't know if you are familiar with Euler's formula, but if you are then those trig formulas are easy to derive:
[tex]e^{ix}=\cos{x}+i*\sin{x}[/tex]
so
[tex]e^{i2x}=(\cos{x}+i*\sin{x})^2[/tex]
[tex]=\cos^2{x}+2i*\sin{x}*\cos{x}-\sin^2{x}[/tex]
Since
[tex]\sin{x}=Im:e^{ix}[/tex]
Then
[tex]\sin{2x}=2*\sin{x}\cos{x}[/tex]
You also get the double angle formula for cosine for free. If you do not know Euler's formula, then you can still prove this geometrically, but it will take more work.

 

[VietDao29]

sin (x+y) = sin x cos y + cos x sin y. Simply place an "x" for the "y" in the formula. Noted: sin (2x) = 2sin x cos x.

?
This is not correct.
What if I say that: sin(x + y) = sin(x)sin(y) + cos(x)cos(y) + sin(x)cos(y) + sin(y)cos(x) - 1.
It certainly satisfies: sin(2x) = sin(x + x) = 2sin(x)cos(x). But it's not true, right?
And moreover, it's some kind of circular argument. One should know the angle sum identities before they know the double identities.
There's a geometry proof at the end of this article. (it works for 0 <= x, y <= 90^o). One can then show that the identity is true for every angle.

 

[d_leet]

?
This is not correct.
What if I say that: sin(x + y) = sin(x)sin(y) + cos(x)cos(y) + sin(x)cos(y) + sin(y)cos(x) - 1.
It certainly satisfies: sin(2x) = sin(x + x) = 2sin(x)cos(x). But it's not true, right?

It's certainly true if x=y but not in general. But sin(2x) does equal 2sin(x)cos(x) because sin(x+y) = sin(x)cos(y) + cos(x)siny(y)