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Hannah_Ahn

  • 4 years ago

prove. sin [(pi over 4)+x] + sin [(pi over 4)-x] = sqrt{2} cosx

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  1. myininaya
    • 4 years ago
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    i think expanding both of those things might help

  2. sam_unleashed
    • 4 years ago
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    use ... sin(a=b)= sin(a)cos(b) +cos(a)sin(b) and sin(a-b)= sin(a)cos(b) -cos(a)sin(b)

  3. Hannah_Ahn
    • 4 years ago
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    all i know is that i will have to use addition indentities.

  4. myininaya
    • 4 years ago
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    \[[\sin(\frac{\pi}{4})\cos(x)+\sin(x)\cos(\frac{\pi}{4})]+[\sin(\frac{\pi}{4})\cos(x)-\sin(x)\cos(\frac{\pi}{4})]\]

  5. myininaya
    • 4 years ago
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    \[\frac{\sqrt{2}}{2} \cos(x)+\sin(x) \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\cos(x)-\sin(x)\frac{\sqrt{2}}{2}\]

  6. myininaya
    • 4 years ago
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    combine like terms and you will be done

  7. sam_unleashed
    • 4 years ago
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    its 2sin(a)cos(b)....now plug valus.....(2/\[\sqrt{2}\] ) sinx

  8. sam_unleashed
    • 4 years ago
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    \[\sqrt{2}\] sin x

  9. Hannah_Ahn
    • 4 years ago
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    you rock myininaya!!! thank you so much!!! and i appreciate your work and help Sam_unleashed! :)

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