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RobZBHayes

  • one year ago

How do you convert sin into cos, and vice versa. Promblem: cos(6x) = sin(3x-9)

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  1. svaibhavi
    • one year ago
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    cos(6x)= sin(90-6x)= sin(3x-9) sin values are equal u can equate the angles 90-6x=3x-9 99=9x x=11

  2. satellite73
    • one year ago
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    i don't think this that easy take a look at the solutions http://www.wolframalpha.com/input/?i=cos%286x%29%3Dsin%283x-9%29

  3. RobZBHayes
    • one year ago
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    I'm not sure if it's with radians or not but I'm in trig if that helps any. I'm just looking for a conversation rate if there is one.

  4. Hero
    • one year ago
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    6x + 3x - 9 = 90

  5. ZeHanz
    • one year ago
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    You want to convert to an equation of the form cos a = cos b or sin a = sin b, because these kind of equations have standard solutions:\[\cos a=\cos b \Leftrightarrow a= \pm b+2k \pi\]\[\sin a = \sin b \Leftrightarrow a=b+2k \pi \vee a=\pi-b+2k \pi\](assuming radians) To get to one of these forms, you can use the following conversions:\[\cos x=\sin(\frac{ \pi }{ 2 }-x)\]or \[\sin x =\cos(\frac{ \pi }{ 2 }-x)\]

  6. Hero
    • one year ago
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    x = 11 sin(66) = sin(24) in degrees

  7. RobZBHayes
    • one year ago
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    @ZeHanz So you're saying I would take cos(6x) and subtract 6x from pi/2 giving me sin(pi/2 - 6x) being equal to cos(6x)

  8. RobZBHayes
    • one year ago
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    @Hero I appreciate the answer but I'm looking for an explanation

  9. Hero
    • one year ago
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    sin(x) = cos(y) are co-functions, meaning angles x and y are complementary.

  10. ZeHanz
    • one year ago
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    @RobZBHayes : exactly!

  11. RobZBHayes
    • one year ago
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    @ZeHanz alright so sin(pi/2 - 6x) = sin(3x-9) and cos(6x) = sin(3x-9) are the same problem, correct? So how would I go about solving sin(pi/2 - 6x) = sin(3x-9)?

  12. Hero
    • one year ago
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    You mean it is the same type of problem, right?

  13. Hero
    • one year ago
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    But not the same exact problem.

  14. Hero
    • one year ago
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    For sin(pi/2 - 6x) = sin(3x - 9) the angles are equal so pi/2 - 6x = 3x - 9

  15. Hero
    • one year ago
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    Solve for x

  16. RobZBHayes
    • one year ago
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    I tried that and came up with an answer that isn't in my answer choices. I got 0.67453292519943295769236907684886, I'm not sure if it's correct or not though. First I added 6x to both sides giving me pi/2=9x-2, then I multiplied everything by 2 giving me pi=18x-9, after that I added 9 to both sides leaving me with 12.14159265358979323846264338328=18x, finally I divided everything by 18 giving me x=0.67453292519943295769236907684886. Did I do all of that correctly? @Hero @ZeHanz

  17. Hero
    • one year ago
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    What are your answer choices?

  18. RobZBHayes
    • one year ago
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    x=10 x=-10 x=11 x=12 You said before that the answer is 11 and I think that's right but I'm trying to get the relation between sin, cos, and tan in case I come across another problem like this one.

  19. Hero
    • one year ago
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    Yes, that's what I got

  20. Hero
    • one year ago
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    pi/2 - 6x = 3x - 9 180/2 - 6x = 3x - 9 90 + 9 = 6x + 3x 99 = 9x 99/9 = x 11 = x

  21. Hero
    • one year ago
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    Remember, pi = 180

  22. Hero
    • one year ago
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    I already explained to you the relationship sin(x) = cos(y) if x and y are complementary tan(x) = cot(y) if x and y are complementary csc(x) = sec(y) if x and y are complementary

  23. Hero
    • one year ago
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    Those are properties of co-functions

  24. RobZBHayes
    • one year ago
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    So the covertion of sin to cos and vice versa is always pi/2-x and pi always equals 180?

  25. Hero
    • one year ago
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    pi = 180 because the length of pi is a semicircle having a measure of 180 degrees.

  26. RobZBHayes
    • one year ago
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    Oh ok, I wasn't aware that pi was a semicircle lol. I guess that's kind of sad considering I'm in trig -_-

  27. ZeHanz
    • one year ago
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    I guess "being in trig" means you have to do everything in degrees and do not need to worry about angles greater than 180 degrees (or 360?) and smaller than 0? In that case I would solve the equation as follows:\[\cos 6x=\sin (3x-9) \Leftrightarrow \cos 6x=\cos(90 - (3x-9))=\cos((99-3x)\]This would give:\[6x=\pm (99-3x)\]which means\[6x=99-3x \vee 6x=3x-99 \Leftrightarrow\]\[9x=99 \vee 3x=-99 \Leftrightarrow x=11 \vee x=-33\]So x=11 would be the only answer that is left, so option C

  28. RobZBHayes
    • one year ago
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    So if I want to solve cos(2x-4)=sin(6x) the first thing I would want to do is convert sin(6x) into a cos, and that would be cos(pi/2-6x). Then my equation would be 2x-4=pi/2-6x, so if I were to simplify that It would be 2x-4=90-6x, then I would want to subtract 2x from both sides giving me -4=90-8x, then I would subtract 90 from both sides giving me -94=-8x, finally I would want to divide everything by -8x giving me my final answer as x=11.75 as a final answer, right?

  29. ZeHanz
    • one year ago
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    Yes, but there is no need to use radians first and then degrees. If you do this, you make things much more difficult for yourself.Either do everything in degrees, or do evertything in radians. Further, I'm still unsure about what "being in trig" means. I would get:\[2x-4= \pm 90-6x\]so\[2x-4=90-6x \vee 2x-4=6x-90 \Leftrightarrow\]\[8x=94 \vee 4x=86 \Leftrightarrow\]\[x=11.75 \vee x=21.5\]The last solution coming from the negative solution, which can probably be discarded...

  30. RobZBHayes
    • one year ago
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    Trig is a shortened way of saying Trigonometry

  31. ZeHanz
    • one year ago
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    Which means every angle is an angle in a triangle?

  32. Hero
    • one year ago
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    If you wanted to solve cos(2x-4)=sin(6x) knowing that angles 2x - 4 and 6x are complementary you would simply add both equations together and set them equal to 90 2x - 4 + 6x = 90 8x - 4 = 90 8x = 94 x = 94/8 x = 11.75 in degrees.

  33. Hero
    • one year ago
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    @ZeHanz, your approach seems to include unnecessary steps.

  34. Hero
    • one year ago
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    All one needs to know is that if sin(x) = cos(y) then you are dealing with co-functions where x and y are complementary.

  35. RobZBHayes
    • one year ago
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    Our of curiosity how would you determine if the angles are in fact complementary in these types of equations?

  36. Hero
    • one year ago
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    If you see a situation where you have an equation of the form sin(x) = cos(y), then x and y must be complementary.

  37. Hero
    • one year ago
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    The equation in that form is the definition of a co-function. That's how you know.

  38. Hero
    • one year ago
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    It's not the mystery you're making it out to be. Co-functions are simply a property of trigonometric functions

  39. Hero
    • one year ago
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    http://www.barstow.edu/lrc/tutorserv/handouts/045%20Trigonometric%20Properties.pdf

  40. Hero
    • one year ago
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    sin(x) = cos(y) is a general form of a co-function identity where x and y are complementary angles such that x + y = 90 sin(6x) = cos(2x - 4) is a specific form of the co-function identity where solving 6x + (2x - 4) = 90 for x reveals the complementary angles.

  41. Hero
    • one year ago
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    It's as simple as that.

  42. ZeHanz
    • one year ago
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    @Hero: I understand now these equations are not about the sin and cos functions defined for every real number, but that they are connected to sin and cos as defined in triangles. That explains my "unnecessary steps" ;)

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