But the answer is a negative!
Let x be a real number satisfying sin ( 2 x ) = 2 5 2 4 .
If the product of all possible values of sin x can be expressed as b a , where a and b are coprime positive integers, find a + b .
The answer is 769.
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2 solutions
Almost correct. It is true that sin x satisfy the equation 6 2 5 k 4 − 6 2 5 k + 1 4 4 = 0 . By Vieta's, the product of all sin x is 144/625. But is it true that all 4 of these values of sin x is in the range of [ − 1 , 1 ] ? How do you know that none of these values of sin x doesn't fall outside the range of [ − 1 , 1 ] ?
actually the equation is 625k^4 - 625k^2 + 144 =0 Sorry , Please do note the mistake in solution !!!! And Could we do this way ? Assume k^4 =t^2 So we ll get quadratic equation 625t^2-625t+144=0 Solving roots we ll get it and hence k^2 and then find values of k to know which of them is in the range . Is there any other method to find out ?
S i n ( 2 X ) = S i n ( 3 6 0 + 2 X ) = 2 4 / 2 5 . . . . . . . . . . . S o 2 ∗ S i n X ∗ C o s X = 2 4 / 2 5 . . . . . . . . . I m p l i e s S i n X ∗ 1 − X 2 = 1 2 / 2 5 . A f t e r s q a r i n g S i n 4 ( X ) − S i n 2 ( X ) + 1 2 2 / 2 5 2 = 0 . S o l v i n g q u a d r a t i c i n S i n 2 ( X ) S i n 2 ( X ) = 1 / 2 ∗ ( 1 ± 1 − 4 ∗ 1 2 2 / 2 5 2 , S i n ( X ) 2 = 9 / 2 5 a n d 1 6 / 2 5 , S o S i n ( X ) = 3 / 5 , 4 / 5 i n 1 s t q u a d r a n t . S i n c e S i n ( 2 X ) = S i n ( 3 6 0 + 2 X ) h a s a r a n g e o f 0 t o 1 8 0 a n d 3 6 0 t o 4 5 0 a n d w i t h f o u r + t i v e a n g l e s , S i n ( X ) t o o w i l l h a v e o t h e r t w o a n g l e s w i t h S i n ( 1 8 0 + X ) = − 3 / 5 , a n d − 4 / 5 . 3 / 5 ∗ 4 / 5 ∗ ( − 3 / 5 ) ∗ ( − 4 / 5 ) = 1 4 4 / 6 2 5 . 1 4 4 + 6 2 5 = 7 6 9 .