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Math Help - Simplifying

  1. #1
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    Simplifying

    If 270^\circ < x < 360^\circ, simplify \sqrt{2+\sqrt{2+2cosx}}

    I couldn't do anything else after getting rid of the square roots..
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  2. #2
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    Quote Originally Posted by Punch View Post
    If 270^\circ < x < 360^\circ, simplify \sqrt{2+\sqrt{2+2cosx}}

    I couldn't do anything else after getting rid of the square roots..
    Use the half angle identity for cosine:

    \cos{\frac{\theta}{2}} = \sqrt{\frac{1 + \cos{\theta}}{2}}.



    \sqrt{2 + 2\cos{x}} = \sqrt{2(1 + \cos{x})}

     = \sqrt{\frac{4(1 + \cos{x})}{2}}

     = 2\sqrt{\frac{1 + \cos{x}}{2}}

     = 2\cos{\frac{x}{2}}.


    Therefore \sqrt{2 + \sqrt{2 + 2\cos{x}}} = \sqrt{2 + 2\cos{\frac{x}{2}}}

     = \sqrt{2\left(1 + \cos{\frac{x}{2}}\right)}

     = \sqrt{\frac{4\left(1 + \cos{\frac{x}{2}}\right)}{2}}

     = 2\sqrt{\frac{1 + \cos{\frac{x}{2}}}{2}}

     = 2\cos{\frac{x}{4}}.
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  3. #3
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    Erm, sorry but the answer in the textbook is 2sin\frac{x}{4}
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  4. #4
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    Then the original question should have been

    \sqrt{2 - \sqrt{2 + 2\cos{x}}}.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Then the original question should have been

    \sqrt{2 - \sqrt{2 + 2\cos{x}}}.
    But i just doubled check both the answer and the question, it is a + for the question
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  6. #6
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    And if the answer provided in the book is wrong, i would like to ask if the step to solving such questions with a square root in another square root is to solve the square roots one by one before combining them all?
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  7. #7
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    Quote Originally Posted by Punch View Post
    And if the answer provided in the book is wrong, i would like to ask if the step to solving such questions with a square root in another square root is to solve the square roots one by one before combining them all?
    Yes, start with the innermost square root and work out...
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  8. #8
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    i think that this question is the toughest question i have met in this chapter though it looks easy
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