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Math Help - Find tan(A - B)

  1. #1
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    Find tan(A - B)

    Find tan(A - B) given the following:

    tanA = 5/12, pi < A < 3pi/2; sinB = -1/2, pi < B < 3pi/2

    NOTE: The red above indicates the given side condition.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by magentarita View Post
    Find tan(A - B) given the following:

    tanA = 5/12, pi < A < 3pi/2; sinB = -1/2, pi < B < 3pi/2

    NOTE: The red above indicates the given side condition.
    two main ways to go about this

    first, you could deduce that B = 210 degrees = \frac {7 \pi}6 radians, and use the identity i gave you below. (you should be able to deduce this based on your knowledge of reference angles and the table that gives you the trig values for special angles, which your professor likely gave you as well)


    alternatively, note that the angles are in the 3rd quadrant, so that the tangents will be positive. so now, we can forget about signs and worry about magnitudes. recall that \text{sine } = \frac {\text{opposite}}{\text{hypotenuse}}. so \sin B = - \frac 12, means we can draw a right-triangle, with an acute angle B, where the side opposite to B has a length of 1 and the hypotenuse has a length of 2. using Pythagoras' theorem, we can find the length of the other side, and use the fact that \text{tangent } = \frac {\text{opposite}}{\text{adjacent}} to find tanB.

    after finding tanB, use the identity:

    \tan(A - B) = \frac {\tan A - \tan B}{1 + \tan A \tan B}
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  3. #3
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    Smile I see...

    Quote Originally Posted by Jhevon View Post
    two main ways to go about this

    first, you could deduce that B = 210 degrees = \frac {7 \pi}6 radians, and use the identity i gave you below. (you should be able to deduce this based on your knowledge of reference angles and the table that gives you the trig values for special angles, which your professor likely gave you as well)


    alternatively, note that the angles are in the 3rd quadrant, so that the tangents will be positive. so now, we can forget about signs and worry about magnitudes. recall that \text{sine } = \frac {\text{opposite}}{\text{hypotenuse}}. so \sin B = - \frac 12, means we can draw a right-triangle, with an acute angle B, where the side opposite to B has a length of 1 and the hypotenuse has a length of 2. using Pythagoras' theorem, we can find the length of the other side, and use the fact that \text{tangent } = \frac {\text{opposite}}{\text{adjacent}} to find tanB.

    after finding tanB, use the identity:

    \tan(A - B) = \frac {\tan A - \tan B}{1 + \tan A \tan B}
    Great reply!
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