Sinx vs. Cosx: What Are the Key Differences?

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What differences is there between a sinx and cosx function?
Do they both use this equation?
y=Asin(BX+C)+D
 
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That's too in depth for me.
I just wanted to know the difference between
y=sinx and y=cosx when graphed.
 
Ok, try this: http://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Graphing%20Sin%20Cos.html
 
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Maybe cristo just missed it, but Ry122 is actually right, cosine can also be expressed in that form of asin(bx+c) + d.

This happens because sin (90degrees-x)= cos x.

So in answer to your original question, the difference when graphed between sin x and cos x is that the cos graph is the same as sins, moved back 90 degrees to the left.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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