Tuesday, July 24, 2012

3sinA + 5cosA = 5 then 3cosA - 5sinA = ?

( a sin A + b cos A )^2 + ( b sin A - a cos A) ^2

= a^2 sin ^2 A + b^2 cos^2 A + 2ab sin A cos A + b^2 sin ^2 A + a^2 cos ^2 A - 2ab cos A sin A
= a^2 + b^2

so
(3sinA + 5cosA)^2 + (5 sin A - 3 cos A)^2 = 3^2 + 5^2

so 5^2 + (5 sin A - 3 cos A)^2 = 3^2 + 5^2

so (5 sin A - 3 cos A) = +/- 3

or 3 cos A - 5 sin A = +/- 3

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