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    Can someone help me with question 6x?

    I don't know how to get the answer as (cos B, sin B) / cos Bi + sin Bj

    I know you have to divide the vector by the magnitude (by using square root) etc

    but I don't know how it turned 1/r = cos B and tan B/r = sin B?

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    What do you get for the modulus? You should be able to factor a common factor out of the sum and identify a trig identity.
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    What do you mean? Don't I have to divide by r ?
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    Apologies for my last post, are we talking about question 10? r is the modulus for question 9, but question 10 has no r in it and so the modulus cannot be r. Perhaps you mixed those two up?
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    (Original post by aznkid66)
    Apologies for my last post, are we talking about question 10? r is the modulus for question 9, but question 10 has no r in it and so the modulus cannot be r. Perhaps you mixed those two up?
    it was question 10, but I thought when you find the unit vectors -you had to divide by a r value. I'm not sure what to divide question 10 by?
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    Find the modulus the good ol' cartesian way. Hint:

    \sin^2 x + \cos^2 x = 1

    \sec ^2 (sin^2x + \cos^2 x )= \sec^2 (1)

    Multiply it through, and you'll find a nice tan(x) identity.
 
 
 
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