STEP II 1990 question 2 solution

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STEP II 1990 question 2 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1990 question 2 solution

$A + B + C + D = \pi \sin A\sin C + \sin B\sin D = \sin (A + B)\sin (A + D) \Leftrightarrow -2\sin A\sin C - 2\sin B\sin D = -2\sin (A + B)\sin (A + D)$

Using the product to sum and compound angle formulae:

$LHS = \cos (A + C) - \cos (A - C) + \cos (B + D) - \cos (B - D) \cos (A + C) = \cos (\pi - (B + D)) = - \cos (B + D) \Rightarrow LHS = - \cos (A - C) - \cos (B - D) = - \cos (C - A) - \cos (B - D)$

This is because cos(x) is an even function.

$- \cos (C - A) = \cos (\pi - (C - A)) = \cos (2A + B + C) \cos (2A + B + D) - \cos (B - D) = -2\sin (A + B) \sin (A + D) = RHS$

For the second part, draw the cyclic quadrilateral with vertices P, Q, R, S, and also the diagonals PR and QS.

Let angle(PRQ) = angle(PSQ) = A, angle(QSR) = angle(QPR) = B, angle(RQS) = angle(RPS) = C, angle(SRP) = angle(SQP) = D. Each pair of angles is equal as they are angles subtended by the same arc.

It is clear that $A + B + C + D = \pi$ , as the angles make up the angles of a triangle.