Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Find the answer

Given that sinA+sinB=1\sin A+\sin B=1 and cosA+cosB=3/2\cos A+\cos B=3 / 2, what is the value of cos(AB)\cos (A-B)?

A number or a short expression. Spacing and $ signs are ignored.

Solution

Squaring both equations and add them together, one obtains 1+9/4=2+2(cos(A)cos(B)+sin(A)sin(B))=2+2cos(AB)1+9 / 4=2+2(\cos (A) \cos (B)+\sin (A) \sin (B))=2+2 \cos (A-B). Thus cosAB=5/8\cos A-B=5 / 8.

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