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Math Help - Help Needed with this Trigonometric Proof

  1. #1
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    Help Needed with this Trigonometric Proof

    Hi, can someone show me how to prove that

    sin A + cos A = (the square root of 2) sin ( A + pi over 4) ?

    Thanks, any help would be very much appreciated!
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  2. #2
    MHF Contributor harish21's Avatar
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    RHS = \sqrt{2}\;\bigg[ \sin\bigg(A+\dfrac{\pi}{4}\bigg)\bigg]

    now use \sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B) where B = \dfrac{\pi}{4}
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  3. #3
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    Hello, sammysparrow!

    \text{Prove that: }\:\sin A + \cos A \:=\:\sqrt{2}\sin\left(A + \frac{\pi}{4}\right)

    Multiply the left side by \frac{\sqrt{2}}{\sqrt{2}}

    . . \frac{\sqrt{2}}{\sqrt{2}}\left(\sin A + \cos A\right) \;=\;\sqrt{2}\left(\sin A\cdot\frac{1}{\sqrt{2}}+ \cos A\cdot\frac{1}{\sqrt{2}}\right) .[1]


    Note that: . \sin\frac{\pi}{4} \,=\,\cos\frac{\pi}{4} \,=\,\frac{1}{\sqrt{2}}

    Then [1] becomes: . \sqrt{2}\left(\sin A\cos\frac{\pi}{4} + \cos A\sin\frac{\pi}{4}\right) \;=\;\sqrt{2}\sin(A +\frac{\pi}{4})

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  4. #4
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    Quote Originally Posted by sammysparrow View Post
    Hi, can someone show me how to prove that

    sin A + cos A = (the square root of 2) sin ( A + pi over 4) ?

    Thanks, any help would be very much appreciated!
    If you had to begin from the LHS, another way is

    2SinACosB=Sin(A+B)+Sin(A-B)

    2CosASinB=Sin(A+B)-Sin(A-B)

    \Rightarrow\ 2(SinACosB+CosASinB)=2Sin(A+B)

    \Rightarrow\ SinACosB+CosASinB=Sin(A+B)


    SinACosB+CosASinB=k(SinA+CosA)

    \Rightarrow\ SinB=CosB\Rightarrow\ B=\frac{\pi}{4}

    \displaystyle\ Sin\frac{\pi}{4}=Cos\frac{\pi}{4}=\frac{1}{\sqrt{2  }}

    \Rightarrow\ SinA+CosA=\sqrt{2}\left[SinA+\frac{\pi}{4}\right]
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  5. #5
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    Thanks everyone!
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