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Determine A, B and C so that the formula (cosx−5sinx)^2 = A cos2x + B sin2x + C is valid for all values on x. A, B och C are integers.

This is translated from a trigonometry problem in another language, I hope you understand it.

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Trigonometry is a language. Is the equation an exact replication?
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1 Answer

after squaring left side you get
cos2x -10sinxcosx +25sin2x or
24sin2x -12 + 12 + cos2x + sin2x - 5sin2x
=12(2sin2x - 1) + 13 - 5 sin2x
=-12(cos22x) - 5 sin 2x + 13, so
A=-12, B = -5, C = 13.
 
Check my work...I hurried to get your answer quickly.

Comments

Could you please explain what you did after squaring the left side. How did you go from cos2x -10sinxcosx +25sin2x to 24sin2x -12 + 12 + cos2x + sin2x - 5sin2x ?
cos 2x = cos2x - sin2x = 1 - 2sin2x = 2 cos2x;
this enables you to get from squared versions into cosine(2x) ... double angle formula for cosine
 
sin 2x = 2 sinx cosx ... double angle formula for sine
 
also, the Pythagorean identity was utilized: sin2x+cos2x=1