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  • 4 years ago

Help with 1a 6b

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  1. anonymous
    • 4 years ago
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    I don't know if this is the fastest way, but here's what I would do: \[A \sin(x+c) = A(sinx cosc + sinccosx)\] Since there already is a sinx - cosx, we know that c will be in 4th quadrant because sinc must be negative and cosc must be positive, and that cosc = sinc (ignoring the signs) This gives us \[3\pi/4\] or \[-\pi/4\] And since we have a common factor of \[\sqrt{2}/2\] we need to make A to cancel that so A will = \[\sqrt{2}\] And so we get: \[\sqrt{2} \times (\sin(x+3\pi/4))\] or \[\sqrt{2} \times (\sin(x-\pi/4))\]

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