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December 20, 2014

December 20, 2014

Posted by **Man** on Sunday, May 25, 2014 at 7:05pm.

Answer: The sine graph will always pass through (0, 0), While the cosine graph wouldn't. If the graph is 2cosx, it will pass through (0, 2). How high the graphs go depend on its amplitude, (how high or how far the graphs go can sometimes be infinite).

Part 2: Using these similarities and differences, how would you transform f(x) = 3 sin(4x - π) + 4 into a cosine function in the form f(x) = a cos(bx - c) + d?

Answer: I don't know, please help

- Algebra II -
**Damon**, Sunday, May 25, 2014 at 7:16pmtrig identity:

sin theta = cos (pi/2-theta)

so

y = 3 sin (4x-pi) + 4

but from above trig identity

y = 3 cos [ pi/2 - (4x-pi) ] + 4

or

y = 3 cos (-4x + 3 pi/2) + 4

but

cos (- theta) = cos theta

so

y = 3 cos (4x-3 pi/2) + 4

- Algebra II -
**Man**, Sunday, May 25, 2014 at 7:18pmThank you

- Algebra II -
**Damon**, Sunday, May 25, 2014 at 7:22pmalso in your explanation explain that the cosine curve is 90 degrees or pi/2 radians behind the sine curve if you graph y = sin theta and y = cos theta versus theta

- Algebra II -
**Damon**, Sunday, May 25, 2014 at 7:22pmYou are welcome :)

- Algebra II -
**Damon**, Sunday, May 25, 2014 at 7:33pmBy the way, there are an infinite number of solutions. eg 3 cos (4x - 7 pi/2)+4

or

- 3 cos(4x -pi/2) + 4

etc

- Algebra II -
**debra**, Sunday, May 25, 2014 at 10:44pm3-x/(x+3)(x-3)

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