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advanced functions

Angle a lies in the second quadrant and sina=7/25.
a) Determine an exact value for cos 2a
b) Determine an exact value for sin 2a

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Anonymous

  1. There must be a 7,24,25 triangle involved.

    cosa=-24/25

    cos2a=sin^2 a - cos^2 a
    sin2a=2sinacosa

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    bobpursley

  2. Before you do anything, start by drawing this angle on the unit circle (this does not have to be to scale). Because angle 𝘒 is in quadrant 2, the domain of the angle must be between πœ‹ and πœ‹/2. Your graph should look like a single line between πœ‹ and πœ‹/2. Now, draw a dotted line from the tip of your line pointing to the left side of your graph down to the x-axis. Draw angle 𝘒 between the x-axis and your line. You should now have a triangle in the second quadrant.

    Now take the information you have for sine to construct your triangle. We know that to find sine, we divide the opposite side of the triangle by its hypotenuse. So, if sin𝘒 = 7/25, then the opposite side of our triangle is 7 and the hypotenuse must be 25. Find the length of the last side by using the pythagorean theorem:

    a^2 + b^2 = c^2
    b^2 = c^2 - a^2
    b = √(25)^2 - (7)^2)
    b = √625 - 49
    b = √576
    b = 24

    Therefore, our triangle has a side of 7 opposite to 𝘒, another side of 24 adjacent to 𝘒, and a hypotenuse of 25. Now we can solve for the next questions.

    a) Use the triangle and the double angle identity to solve for cos2𝘒:

    (Remember, because we are in quadrant 2, the CAST rule says cos is negative, so we need to indicate that at the start. Sine however is positive, so we can leave that as it is.)
    -cos2𝘒 = -cos^2𝘒 - sin^2𝘒
    = (-24/25)^2 - (7/25)^2
    ↑ Because -24/25 is squared, the negative sign disappears
    = (576/625) - (49/625)
    = 527/625

    Therefore, cos2𝘒 is 527/625.

    b) Use the triangle and the double angle identity to solve for sin2𝘒:

    sin2𝘒 = 2sin𝘒 cos𝘒
    = 2(7/25)(-24/25)
    ↑ Remember cos is negative
    = 2(-168/625)
    = -336/625

    Therefore, sin2𝘒 is -336/625.

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    Anonymous

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