trig p5 Watch

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ryan750
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#1
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#1
solve:

5cosx - 12sinx = 6.5
for 0<x<360

thanks
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RichE
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#2
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#2
Use [email protected][email protected] = Rcos(@-c)

where

R^2 = A^2 + B^2

and

tanc = B/A
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ryan750
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(Original post by RichE)
Use [email protected][email protected] = Rcos(@-c)

where

R^2 = A^2 + B^2

and

tanc = B/A
did that - my answer doesnt agrree with book - i realy cant se where im going wrong

hold on i got it. Why did u use - c and not + c
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RichE
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(Original post by ryan750)
solve:

5cosx - 12sinx = 6.5
for 0<x<360

thanks
R^2 = A^2 + B^2 = 25 + 12^2 = 13^2 -> R =13

tanc = -12/5

c = -arctan(12/5) = -1.176

13cos(@+1.176) = 6.5

cos(@+1.176) = 0.5

@+1.176 = 5.2359 or 7.3303

@ = 4.06 or 6.154
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ryan750
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#5
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#5
no its stil comin out rong - can sum1 see what they get
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ryan750
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#6
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ok - i get same answers as u. But i dont see why u have put - c instead of + c.
i thought that when u have a positive cosine and negative sine u use
cos(x + c) because when all terms positive u use cos(x - c).

But the answers we have differ to the book. It has 20.3 and 260.3 degrees?

i still dont get to this if i use cos(x+c)? maybe book is wrong.
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RichE
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(Original post by ryan750)
ok - i get same answers as u. But i dont see why u have put - c instead of + c.
i thought that when u have a positive cosine and negative sine u use
cos(x + c) because when all terms positive u use cos(x - c).

But the answers we have differ to the book. It has 20.3 and 260.3 degrees?

i still dont get to this if i use cos(x+c)? maybe book is wrong.

It doesn't matter which you use as long as you are careful in equating the coefficients of [email protected] and [email protected] when expanding either
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RichE
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(Original post by ryan750)
But the answers we have differ to the book. It has 20.3 and 260.3 degrees?

i still dont get to this if i use cos(x+c)? maybe book is wrong.
you can check whether the book is wrong by putting its answers back into the equation
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ryan750
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#9
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#9
book is rong - nit all
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