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the y-zeroes are at 0, π, 2π etc: what is tan(2*0), tan(2*π), etc what is cos(4*0), cos(4*π), etc what is sin(1*0), sin(1*π), etc the curves peak/ trough at 1/-1, and are symmetrical about the x-axis. that must count for something too. they have not been shifted around. you can work this out from this and any other clues you might find. cos(x-π)!! that is interesting too. that shifts the cos curve back by π on the x-axis. however the final option also shifts the curve on the y-axis as it is "-2-cos(x-pi)". that means, surely, that the y values must be between -1 and -2?!?!
is it C?
Test some values. 0 is a good one. Does 0 in the formula for C give you the right y-value, judging from the graph?
You know the graph is continuous, so that excludes. Then you know that the range is +- 1, so the amplitude is 1 and the amplitude of cos and sin is1, so that excludes... which one. Finally you know the graph is is centered at y = 0, so that excludes ...? Then you have one left.
As a favorite math teacher used to say whenever someone offered up an answer like "is it C?" — "is that an answer, or a prayer?" :-)
if x=0 y should = 0 too right? based on the graph
if you think the answer is C then try insert x = 0 Sin(0+pi) = ? sin(0) = 0, sin(pi/2) = 1 sin(pi) = ?
C is correct. We've ruled out A because the graph of tan doesn't look like a sine curve. We can rule out B because the amplitude of the curve is wrong - it goes from -3 to 3 instead of -1 to 1 like our graph. D can be ruled out because it has an offset of -2 and even without the offset, the phase is wrong, because cos(x-pi) at x = 0 is -1, and our curve needs to go through (0,0). That leaves C, y = sin(x+pi). At x = 0, sin(0+pi) = sin(0) = 0, so that matches our graph. At x = pi, sin(pi+pi) = 0, and that also matches our graph. Not only is it the correct answer by process of elimination, it actually matches :-)