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    Q: express 3sin+4cos in the form Rsin(theata + alpha). And 0< alpha < 90 degrees
    (!)= theata
    (?)= alpha


    ii) hence, solve the equation 3sin(!) + 4cos(!) +1 = 0

    solutions between -180 and 180

    Ive done part1 which gives the angle 53.1.....

    the he second part I'm stuck with as I don't think I'm correct, here is my working out for part 2

    3sin(!) +4cos(!) = inverse tan(4/3)

    therefore re can I just make it

    tan = 4/3 -1

    inverse tan = 18.4349.....


    i have no clue what I'm doing thus me guessing what to do...
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    (Original post by ADotCross)
    Q: express 3sin+4cos in the form Rsin(theata + alpha). And 0< alpha < 90 degrees
    (!)= theata
    (?)= alpha


    ii) hence, solve the equation 3sin(!) + 4cos(!) +1 = 0

    solutions between -180 and 180

    Ive done part1 which gives the angle 53.1.....

    the he second part I'm stuck with as I don't think I'm correct, here is my working out for part 2

    3sin(!) +4cos(!) = inverse tan(4/3)

    therefore re can I just make it

    tan = 4/3 -1

    inverse tan = 18.4349.....


    i have no clue what I'm doing thus me guessing what to do...
    have you got any notes from school?
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    Yes, just flicking through them
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    (Original post by ADotCross)
    Q: express 3sin+4cos in the form Rsin(theata + alpha). And 0< alpha < 90 degrees
    (!)= theata
    (?)= alpha


    ii) hence, solve the equation 3sin(!) + 4cos(!) +1 = 0

    solutions between -180 and 180

    Ive done part1 which gives the angle 53.1.....

    the he second part I'm stuck with as I don't think I'm correct, here is my working out for part 2

    3sin(!) +4cos(!) = inverse tan(4/3)

    therefore re can I just make it

    tan = 4/3 -1

    inverse tan = 18.4349.....


    i have no clue what I'm doing thus me guessing what to do...
    Try using your answer from part i) to solve part ii) so..

    Rsin(θ + 53.1) = -1 (using the value you calculated for R)
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    So it would be

    inverse tan-1/5 - 53.1.... Then makes missing theata number ?
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    (Original post by ADotCross)
    So it would be

    inverse tan-1/5 - 53.1.... Then makes missing theata number ?
    Why do you keep doing "inverse tan"?

    You have an equation that is basically

    sin(something) = something else

    so the first thing you should be doing is an inverse sine!
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    Sin(theata +53.1) = -1/5

    can I then inverse sin to make

    theata + 53.1 = inverse sin(-1/5)

    then - 53.1

    theata = -64.63...

    is is that how you do it ?
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    (Original post by ADotCross)
    Sin(theata +53.1) = -1/5

    can I then inverse sin to make

    theata + 53.1 = inverse sin(-1/5)

    then - 53.1

    theata = -64.63...

    is is that how you do it ?
    Eseentially, although you'll need to do something to get angles within the range of values they've asked for.

    (and it's theta by the way )
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    Yh I got the rest thanks, I found the angles between the range.
 
 
 
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