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Math Help - sine help

  1. #1
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    sine help

    Hello all... just took a test in my pre calc class and got it back with a few wrong. We can correct them and then retake the test but im having trouble with corrections.

    These are the ones that im having trouble with:

    solve the equation -

    2cos(πx + π/3) = - sqrt(2)

    sin2x = sin6x

    sin2x = -sqrt(2) * sin x

    Im aware of the formulas and ive tried using them for these but am not getting any success.

    for the first one i used arccosine after dividing each side by 2.. easy solve after that... got 5/12 and 11/12 but was marked wrong

    for the second one i was not to sure but i said 0 was the answer.

    for the last one i tried dividing by sin x and then using the formula sine2x = 2 sine x cos x... then turning the left side to 2sin x tan x. After dividing by 2 you can solve but still got the wrong answers...

    Help would be a great thing and thanks in advance for anyone who has any suggestions
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by rm1991 View Post
    Hello all... just took a test in my pre calc class and got it back with a few wrong. We can correct them and then retake the test but im having trouble with corrections.

    These are the ones that im having trouble with:

    solve the equation -

    2cos(πx + π/3) = - sqrt(2)

    sin2x = sin6x

    sin2x = -sqrt(2) * sin x

    Im aware of the formulas and ive tried using them for these but am not getting any success.

    for the first one i used arccosine after dividing each side by 2.. easy solve after that... got 5/12 and 11/12 but was marked wrong

    for the second one i was not to sure but i said 0 was the answer.

    for the last one i tried dividing by sin x and then using the formula sine2x = 2 sine x cos x... then turning the left side to 2sin x tan x. After dividing by 2 you can solve but still got the wrong answers...

    Help would be a great thing and thanks in advance for anyone who has any suggestions
    for the last one 2sin(x)cos(x)=-\sqrt{2}sin(x)\Rightarrow{2sin(x)cos(x)+\sqrt{2}si  n(x)=0} \Rightarrow{sin(x)(2cos(x)+\sqrt{2})=0} now use the zero product property
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  3. #3
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    thanks i get it... didnt see it like that at first
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by rm1991 View Post
    Hello all... just took a test in my pre calc class and got it back with a few wrong. We can correct them and then retake the test but im having trouble with corrections.

    These are the ones that im having trouble with:

    solve the equation -

    2cos(πx + π/3) = - sqrt(2)

    sin2x = sin6x

    sin2x = -sqrt(2) * sin x

    Im aware of the formulas and ive tried using them for these but am not getting any success.

    for the first one i used arccosine after dividing each side by 2.. easy solve after that... got 5/12 and 11/12 but was marked wrong

    for the second one i was not to sure but i said 0 was the answer.

    for the last one i tried dividing by sin x and then using the formula sine2x = 2 sine x cos x... then turning the left side to 2sin x tan x. After dividing by 2 you can solve but still got the wrong answers...

    Help would be a great thing and thanks in advance for anyone who has any suggestions
    For the second one I would use sin(6x)=sin(2x+4x) then go from there..and 0 would be an answer
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  5. #5
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    hmmm makes since! thanks a lot

    oh btw, there is another one about a triangle that i thought i had but seem to be lost with

    We are not allowed to use a calculator to make calculations with this and he gave us an approximate value of 0.6 for cos 53... We were supposed to find the area of a triangle with one side 35 another side 4 and angle between them 53.

    I tried using the law of cosines and herons theorem but theres a sqrt after you use law of cosines to find the other side... so i would assume that this is not the way to do it (since were not supposed to use a calculator to find the answer)
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by rm1991 View Post
    hmmm makes since! thanks a lot

    oh btw, there is another one about a triangle that i thought i had but seem to be lost with

    We are not allowed to use a calculator to make calculations with this and he gave us an approximate value of 0.6 for cos 53... We were supposed to find the area of a triangle with one side 35 another side 4 and angle between them 53.

    I tried using the law of cosines and herons theorem but theres a sqrt after you use law of cosines to find the other side... so i would assume that this is not the way to do it (since were not supposed to use a calculator to find the answer)
    C^2=16+35^2-2\cdot{4}\cdot{3}5\cdot{.6}...group and factor and its not that difficult...Heron's Formula...with this ! I am assuming it was multiple choice...regardless the operative word in the directions is...approximately
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  7. #7
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    wait but the law is  c^2 = a^2 + b^2 - 2abCosC ... so where did the minus 6 come from? and no.. not multiple choice sadly
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by rm1991 View Post
    wait but the law is  c^2 = a^2 + b^2 - 2abCosC ... so where did the minus 6 come from? and no.. not multiple choice sadly
    Aha! Check again...my typo was fixed...I am goood ...haha just kidding
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  9. #9
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    hmm... ok

    same as i got earlier but dont know how my teacher expects us how to do this with out a calculator... the computations done with heron's theorem must be hellish!
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by rm1991 View Post
    hmm... ok

    same as i got earlier but dont know how my teacher expects us how to do this with out a calculator... the computations done with heron's theorem must be hellish!
    Yes...they scare me...
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  11. #11
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    by the way, for  sin2x = sin 6x i get  sin2x = sin6x \rightarrow sin2x = sin(2x + 4x) \rightarrow sin2x = sin2x * cos4x + cos2x * sin4x and im stuck there
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  12. #12
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    cos 53=0.6
    sin 53=sqrt(1-0.6^2)
    =0.8
    area=0.5*ab sin 53
    =0.5*35*4*0.8
    =56
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