anonymous
  • anonymous
How can you explain this using the unit circle definitions of sine and cosine? How can you explain it using the right triangle definitions of sine and cosine?
Mathematics
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anonymous
  • anonymous
How can you explain this using the unit circle definitions of sine and cosine? How can you explain it using the right triangle definitions of sine and cosine?
Mathematics
chestercat
  • chestercat
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hoblos
  • hoblos
explain what??
anonymous
  • anonymous
You know that for any , neither sin nor cos can be greater than 1. How can you explain this using the unit circle definitions of sine and cosine? How can you explain it using the right triangle definitions of sine and cosine?
anonymous
  • anonymous
sin is opposite/hypotenuse Surely opposite cannot be equal to hypotenuse nor be greater than hypotenuse. So they can't be greater than 1 Same explanation for cos also

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anonymous
  • anonymous
thanks harry potter :)
anonymous
  • anonymous
|dw:1327332163277:dw| ok i need u to imagine, we start from 3 o clock. when it is 3 oclock, the h(hypotenuse) has the length of maximum(1) same as adjacent(not necessary for sine) but the opposite is 0. When it is 2 oclock, the hypotenuse shorten a bit and the opposite increase. When it reach to 12 o clock, hypotenuse is same as opposite. 3 o clock : sin(0/1) 0 2 o clock : sin(...will be quite long to write it) 12 o clock :sin(1/1) = 90
anonymous
  • anonymous
better way to explain it thanks hei

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