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Math Help - solve trig equation help.

  1. #1
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    solve trig equation help.

    Solve, in the interval 0 ≤ θ<360, the following equations. Give answers to the nearest 0.1.

     sin \theta + cos \theta = 1
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  2. #2
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    Hello Tweety
    Quote Originally Posted by Tweety View Post
    Solve, in the interval 0 ≤ θ<360, the following equations. Give answers to the nearest 0.1.

     sin \theta + cos \theta = 1
    Use the fact that \sin 45^o=\cos 45^o = \frac{1}{\sqrt2} and \sin(A+B) = \sin A \cos B + \cos A \sin B to write
    \sin\theta+\cos\theta =\sqrt2 \sin(\theta + 45^o)=1

    \Rightarrow \sin(\theta + 45^o) = \frac{1}{\sqrt2}
    Can you complete it now?

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello TweetyUse the fact that \sin 45^o=\cos 45^o = \frac{1}{\sqrt2} and \sin(A+B) = \sin A \cos B + \cos A \sin B to write
    \sin\theta+\cos\theta =\sqrt2 \sin(\theta + 45^o)=1

    \Rightarrow \sin(\theta + 45^o) = \frac{1}{\sqrt2}
    Can you complete it now?

    Grandad
    Yes thanks,

    But how did you know you had to use  \frac{1}{\sqrt{2}} ?
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  4. #4
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    Hello Tweety
    Quote Originally Posted by Tweety View Post
    Yes thanks,

    But how did you know you had to use  \frac{1}{\sqrt{2}} ?
    I knew because I've done it so often! But the general method is:

    Let \sin\theta + \cos\theta = R\sin(\theta +A) = R\sin\theta\cos A + R\cos\theta\sin A.

    Then, comparing coefficients of \sin\theta and \cos\theta:
    1 = R\cos A and 1 = R\sin A
    If we now divide the second of these equations by the first, we get:
    \frac{R\sin A}{R\cos A}=\tan A = 1

    \Rightarrow A = 45^o
    And if we square and add the two equations:
    1^2 + 1^2  = R^2(\cos^2A + \sin^2A)

    \Rightarrow R^2=2, since \cos^2A + \sin^2A=1

    \Rightarrow R = \sqrt2
    So there we have it:
    \sin\theta + \cos\theta = \sqrt2\sin(\theta+45^o)
    Grandad
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