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explain using linear approximations, why sin(x)≈x for x near 0
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perhaps you have seen that \[\lim_{x->0}\frac{\sin(x)}{x}=1\]that says the same thing
or you can use "linear approximation" aka find the equation for the line tangent to \[\sin(x)\] at (0,0) the slope is the derivative evaluated at 0. the derivative of sine is cosine, and cosine of 0 = 1 so the line has slope 1
and since it goes through (0,0) the equation for the line i s \[y=x\] so that is the linear approximation for \[\sin(x)\] near x = 0
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