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Thread: 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.

    $\displaystyle 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.

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

    $\displaystyle \Rightarrow \sin(\theta + 45^o) = \frac{1}{\sqrt2}$
    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 $\displaystyle \sin 45^o=\cos 45^o = \frac{1}{\sqrt2}$ and $\displaystyle \sin(A+B) = \sin A \cos B + \cos A \sin B$ to write
    $\displaystyle \sin\theta+\cos\theta =\sqrt2 \sin(\theta + 45^o)=1$

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

    Grandad
    Yes thanks,

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

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

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

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

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

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