A challenging problem indeed!
If x = a cos 2 θ + b sin 2 θ , then find the value of ( x − a ) ( b − x ) − ( a − b ) 2 sin 2 θ cos 2 θ
The answer is 0.
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2 solutions
Thanks Mohamed. The question was originally to prove that ( x − a ) ( x − b ) = ( a − b ) 2 sin 2 θ cos 2 θ
( x − a ) ( b − x ) − ( a − b ) 2 sin 2 θ cos 2 θ = ( a cos 2 θ + b sin 2 θ − a ) ( b − a cos 2 θ − b sin 2 θ ) − ( a − b ) 2 sin 2 θ cos 2 θ = [ a ( cos 2 θ − 1 ) + b sin 2 θ ] [ b ( 1 − sin 2 θ ) − a cos 2 θ ] − ( a − b ) 2 sin 2 θ cos 2 θ = [ − a ( 1 − cos 2 θ ) + b sin 2 θ ] [ b ( cos 2 θ ) − a cos 2 θ ] − ( a − b ) 2 sin 2 θ cos 2 θ = ( − a sin 2 θ + b s i n 2 θ ) ( b cos 2 θ − a cos 2 θ ) − ( a − b ) 2 sin 2 θ cos 2 θ = ( b sin 2 θ − a sin 2 θ ) ( b cos 2 θ − a cos 2 θ ) − ( a − b ) 2 sin 2 θ cos 2 θ = sin 2 θ ( b − a ) cos 2 θ ( b − a ) − ( a − b ) 2 sin 2 θ cos 2 θ = ( a − b ) 2 sin 2 θ cos 2 θ − ( a − b ) 2 sin 2 θ cos 2 θ = 0
It is so easy. By numbers,forexample: the angle=45^,a=2,b=4 So x=3 . (3-2) (4-3) - (2-4)(2-4) × 0.5 × 0.5 = 0 Or any angle, any a ,any b !