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Hannah_Ahn
 3 years ago
simplify:
sin[(3pi over 2)+x)]
Hannah_Ahn
 3 years ago
simplify: sin[(3pi over 2)+x)]

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malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.1\[\sin(\frac{3 \pi}{2}+x)?\]

malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.1\[\sin \left(\frac{3 \pi}{2}+x \right)=\cos(x)\] Because it shifts the sine curve to the LEFT by 3pi/2, which gives you a minimum of 1 at the x=0 which means its the cosine curve, just rotated over the xaxis, i.e., cos(x)

Hannah_Ahn
 3 years ago
Best ResponseYou've already chosen the best response.0what about simplifying: cos (pi2x)?

malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.1I mean, I don't know exactly what it means by simplify, I'm just rewriting the trig function in terms of the other. But if you wanted to do that one then you would take cos(2x) and shift it left or right by pi (since cosine is an even function its symmetric about the y axis). So doing that we get: \[\cos(\pi2x)=\cos(2x\pi)=\cos(2x)\]

malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.1Look at these plots on wolfram if you need to see what I'm explaining.

Hannah_Ahn
 3 years ago
Best ResponseYou've already chosen the best response.0very interesting. i don't think i never learned it that way yet.

Hannah_Ahn
 3 years ago
Best ResponseYou've already chosen the best response.0thanks for your help.

malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.1Well, all you need to know is the transformations. If you have: \[f(ax)\] Make shortens the function in the xdirection by a factor of a. Ex. cos(x) has period 2pi, cos(2x) has period pi (which is why the above problem comes out how it does) http://www.wolframalpha.com/input/?i=plot+y%3Dcos(x)%2Cy%3Dcos(2x) If you have: \[af(x)\] Then that increases the height of the function by a factor of a. So for trig functions (sine and cosine at least) this changes the amplitude. For example: cos(x) has an AMPLITUDE of 1 because it only goes from 1 to 1 (in physics amplitudes are generally from the zero point to a peak or trough) but 2cos(x) has amplitude 2 because it stretches it vertically by 2. http://www.wolframalpha.com/input/?i=plot+y%3Dcos(x)%2Cy%3D2cos(x) If you have: \[f(x+a)\] Then it shifts the function to the LEFT by a (assuming a positive) if a is negative then its shifts it to the RIGHT (yes, its backwards for this one) For example: cos(x) starts at (0,1) (looking at x=0), but cos(x+pi) (as we just saw) starts at (0,1) because it shifts by pi (moving a peak to a trough) http://www.wolframalpha.com/input/?i=plot+y%3Dcos(x)%2Cy%3Dcos(x%2Bpi) If you have: \[f(x)+a\] If a is positive it shifts the function UP by a, and if its negative down by a. For example cos(x) has a y intercept of (0,1) but cos(x)+2 has a y intercept at (0,3) http://www.wolframalpha.com/input/?i=plot+y%3Dcos%28x%29%2Cy%3Dcos%28x%29%2B2 If you have: \[f(x)\] It rotates the function around the x axis. For example, 2cos(x) has a yintercept of (0,2) but 2cos(x) has a yintercept of (0,2) http://www.wolframalpha.com/input/?i=plot+y%3D2cos%28x%29%2Cy%3D2cos%28x%29 And finally, if you have: \[f(x)\] Then it rotates it around the yaxis. For example, if you have sin(x) at x=pi/2 you have (pi/2,1) and it INCREASES through the origin, but sin(x) makes it do the exact opposite (just see link) http://www.wolframalpha.com/input/?i=plot+y%3Dsin%28x%29%2Cy%3Dsin%28x%29 Those are all the transformations (there are rotations and things too but thats different.

malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.1Basically, if you assume you have an f(x) and then you change it in any way I wrote above, those are what happen.

Hannah_Ahn
 3 years ago
Best ResponseYou've already chosen the best response.0Mammamia! You are so awesome :') Thanks so much!

Hannah_Ahn
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1326178415861:dw

Hannah_Ahn
 3 years ago
Best ResponseYou've already chosen the best response.0is it odd/even identities?

malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.1Well, an odd function is one that is symmetric about the origin. Which means, if you rotate it around the xaxis and then the yaxis, you get the same graph. (the sine function) Or: \[f(x)=f(x) \forall x \in \mathbb{R}\] Which means for all real numbers x. If its an even function, that means its symmetric about the xaxis (like the cosine function). Or: \[f(x)=f(x) \forall x \in \mathbb{R}\]

malevolence19
 3 years ago
Best ResponseYou've already chosen the best response.1The properties help SIGNIFICANTLY when you get to classes such as calculus and you can use these to your advantage to simplify overlycomplex problems based on symmetry.
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