anonymous
  • anonymous
simplify: sin[(3pi over 2)+x)]
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
\[\sin(\frac{3 \pi}{2}+x)?\]
anonymous
  • anonymous
yes :)
anonymous
  • anonymous
\[\sin \left(\frac{3 \pi}{2}+x \right)=-\cos(x)\] 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 x-axis, i.e., -cos(x)

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anonymous
  • anonymous
what about simplifying: cos (pi-2x)?
anonymous
  • anonymous
I 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: \[\cos(\pi-2x)=\cos(2x-\pi)=-\cos(2x)\]
anonymous
  • anonymous
Look at these plots on wolfram if you need to see what I'm explaining.
anonymous
  • anonymous
very interesting. i don't think i never learned it that way yet.
anonymous
  • anonymous
thanks for your help.
anonymous
  • anonymous
Well, all you need to know is the transformations. If you have: \[f(ax)\] Make shortens the function in the x-direction 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 y-intercept of (0,2) but -2cos(x) has a y-intercept of (0,-2) http://www.wolframalpha.com/input/?i=plot+y%3D2cos%28x%29%2Cy%3D-2cos%28x%29 And finally, if you have: \[f(-x)\] Then it rotates it around the y-axis. 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%28-x%29 Those are all the transformations (there are rotations and things too but thats different.
anonymous
  • anonymous
Basically, if you assume you have an f(x) and then you change it in any way I wrote above, those are what happen.
anonymous
  • anonymous
Mammamia! You are so awesome :') Thanks so much!
anonymous
  • anonymous
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anonymous
  • anonymous
is it odd/even identities?
anonymous
  • anonymous
Well, an odd function is one that is symmetric about the origin. Which means, if you rotate it around the x-axis and then the y-axis, 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 x-axis (like the cosine function). Or: \[f(-x)=f(x) \forall x \in \mathbb{R}\]
anonymous
  • anonymous
The properties help SIGNIFICANTLY when you get to classes such as calculus and you can use these to your advantage to simplify overly-complex problems based on symmetry.

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