Prove this identity trigonometry?

(cos A - cos B) / (sin A + sin B) = (sin B - sin A) / (cos A + cos B)

4 Answers

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  • TomV
    Lv 7
    2 years ago
    Favorite Answer

    (cos A - cos B) / (sin A + sin B) = (sin B - sin A) / (cos A + cos B)

    LHS = (cosA - cosB)/(sinA + sinB)

    {multiply by conjugate of numerator}

    = (cos²A - cos²B)/[(sinA + sinB)(cosA + cosB)]

    {apply Pythagorean Theorem to numerator and simplify}

    = (sin²B - sin²A)/[(sinA + sinB)(cosA + cosB)]

    {multiply by sinA - sinB}

    = (sin²B - sin²A)(sinA - sinB)/[(sin²A - sin²B)(cosA + cosB)

    = -(sin²A - sin²B)(sinA - sinB)/[(sin²A - sin²B)(cosA + cosB)

    {cancel common term (sin²A - sin²B)}

    = -(sinA - sinB)/(cosA + cosB)

    {distribute negative}

    = (sinB - sinA)/(cosA + cosB)

    = RHS

    QED

    • TomV
      Lv 7
      2 years agoReport

      The "multiply by sinA -sinB" step could more efficiently be accomplished by factoring (sin²B - sin²A) into (sinB + sinA)(sinB - sinA). 20-20 hindsight.

  • cidyah
    Lv 7
    2 years ago

    (cos A - cos B) / (sin A + sin B) = - 2 sin ((A+B)/2) sin ( (A-B)/2) / ( 2 sin ((A+B)/2) cos ((A-B)/2))

    = - sin ((A-B)/2) / cos((A-B))/2)

    = - tan ((A-B)/2)

    (sin B - sin A) / (cos A + cos B) = 2 cos ((A+B)/2) sin ((B-A)/2) / (2 cos ((A+B)/2) cos ((A-B)/2)

    = sin ((B-A)/2) / 3cos ((A-B)/2)

    = - sin ((A-B)/2) / cos((A-B)/2)

    = - tan((A-B)/2)

    Left-hand side = Right-hand side

  • 2 years ago

    (cos A - cos B) / (sin A + sin B) = (sin B - sin A) / (cos A + cos B)

    Cross-multiplying, we get

    (cos A - cos B)*(cos A + cos B) = (sin B - sin A)*(sin A + sin B)

    or (cos^2 A - cos^2 B) = (sin^2 B - sin^2 A)

    or (cos^2 A + sin^2 A) = (sin^2 B + cos^2 B)

    or 1 = 1

  • 2 years ago

    Multiplying to remove the denominators we have:

    (cosA - cosB)(cosA + cosB) = (sinB - sinA)(sinA + sinB)

    Expanding we get:

    cos²A - cos²B = sin²B - sin²A

    Now, sin²A => 1 - cos²A and sin²B => 1 - cos²B

    i.e. cos²A - cos²B = 1 - cos²B - (1 - cos²A)

    => cos²A - cos²B = 1 - cos²B - 1 + cos²A

    or, cos²A - cos²B = -cos²B + cos²A

    so, LHS => RHS

    :)>

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