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anonymous

  • one year ago

The exact value of cos(pi/4) is 1/square root 2 but when working out cos (-7pi/4) why is the answer square root 2/2 ?

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  1. Nnesha
    • one year ago
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    cos is an even function \[\huge\rm Cos(-x)=\cos(x)\] what is cos at 7pi/4 radi ??

  2. anonymous
    • one year ago
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    I have no idea, can you explain it to me further. I got pi/4

  3. Nnesha
    • one year ago
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    so how did you get pi/4 ?? :=)

  4. Nnesha
    • one year ago
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    r u allowed to use unit circle ?:=)

  5. anonymous
    • one year ago
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    by going around the unit circle a lot ?? yes i am

  6. Nnesha
    • one year ago
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    alright

  7. Nnesha
    • one year ago
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    |dw:1440591648001:dw| (x,y) solution where c-coordinate represent cos and y-coordinate = sin so what is cos at 7pi/4 ?

  8. Nnesha
    • one year ago
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    and no pi/4 isn't \[\huge\rm \frac{ 1 }{ \sqrt{2} }\]you can't leave the root at the denominator you have to multiply both top and bottom of the fraction by square root 2

  9. anonymous
    • one year ago
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    how do i find the value of these coordinates if i am not given the unit circle ??

  10. anonymous
    • one year ago
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    oh! so i just have to make sure there are no square roots at the denominator ??

  11. anonymous
    • one year ago
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    and i will get the coordinates i am looking for? thanks!!

  12. Nnesha
    • one year ago
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    yes right \[\textrm {no square root at the denominator }\]

  13. Nnesha
    • one year ago
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    without looking at the unit circle 2 ways 1) familiar with the 30-60-90 and 45-45-90 triangle 2nd) memorize

  14. Nnesha
    • one year ago
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    and for example if they ask `what is the exact value of cos(5pi/4) then cos equal to -sqrt{2} over 2 |dw:1440591987978:dw|

  15. anonymous
    • one year ago
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    because SACT right ?? thank you so much!!

  16. anonymous
    • one year ago
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    **ASTC

  17. Nnesha
    • one year ago
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    ahaha my teacher taught us CAST which i assume same thing ASTC

  18. Nnesha
    • one year ago
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    as*

  19. Nnesha
    • one year ago
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    https://www.youtube.com/watch?v=LE6dmczMc68 here is a video which helps to memorize but i'll show how to find solution by using 45-45-90 theorem just in case if u don't allowed to use unit circle |dw:1440592694816:dw| if you understand the first quadrant thats mean you know all quadrants of the unit circle red liines are increasing by 45 degrees but the blue one increased by 30 degrees

  20. Nnesha
    • one year ago
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    and you know radius of the unit circle is one

  21. Nnesha
    • one year ago
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    |dw:1440592827706:dw| we need to make right angle on the x-axis (always!)

  22. Nnesha
    • one year ago
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    |dw:1440592949487:dw| 45-45-90 is an isosceles triangle which means two sides are identical

  23. anonymous
    • one year ago
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    using pythagoras to find the angles and sides ??

  24. Nnesha
    • one year ago
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    now you can apply the Pythagorean theorem to find value of x \[\huge\rm a^2+b^2=c^2\] c=hypotenuse substitute a, b ,c for their values solve for x you will get the solution thats on the unit circle at 45 degree

  25. Nnesha
    • one year ago
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    yes right!

  26. Nnesha
    • one year ago
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    you already know the angles just need to find solutions (x,y) coordinate

  27. anonymous
    • one year ago
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    awesome thanks! i understand now c: my test is tomorrow hahahahahhahahahahaha

  28. Nnesha
    • one year ago
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    good luck!

  29. phi
    • one year ago
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    FYI \[ \frac{1}{\sqrt{2}}= \frac{2}{\sqrt{2} }\] (use a calculator to see this, or multiply the first fraction by sqr(2)/sqr(2) ) In the "old days", people really did not like square roots in the denominator (too hard to calculate, I think), so they made a point of rationalizing them. These days people still do that, but technically it's not wrong to leave the answer 1/sqr(2)

  30. Nnesha
    • one year ago
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    Well. if you leave the answer as 1/sqrt{2} on the test i'm pretty sure you will get -1

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