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Or perhaps sin 3x = 3 sin x - 4 sin3x. Equate 3 sin x - 4 sin3x to 0 and we have sin x= 0 or 23. Similarly, for sin 3x =0, the basic value for x to satisfy would be 3π. This means sin 3π = 23 and sin−1 (23) =3π . Now, cos 3π = 21 since sin2x + cos2x =1. It follows that cos−1(21)= 3π
The honest truth is YOU CAN'T. First of all, arccos is not an algebraic function, and if you want to make it algebraic you will have to turn it into an infinite power series. The issue then becomes how to determine that arccos(21) is EXACTLY 3π (key word being EXACT, because in algebra we don't have any notion of "convergence", whatever that means).
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Maybe first start with cos 3x = 4 cos3 x - 3 cos x?
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Or perhaps sin 3x = 3 sin x - 4 sin3x. Equate 3 sin x - 4 sin3x to 0 and we have sin x= 0 or 23. Similarly, for sin 3x =0, the basic value for x to satisfy would be 3π. This means sin 3π = 23 and sin−1 (23) =3π . Now, cos 3π = 21 since sin2x + cos2x =1. It follows that cos−1(21)= 3π
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Someone think of a shorter way?
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For 4 cos3 x - 3 cos x=0, we similarly have cos x =0 or 23.
cos 3x = 0 means the smallest positive value for 3x is 2π since cos 2π = 0 which follows that x= 6π.
We can also prove here that cos 6π = 23. Then cos−1(23 ) = 6π.
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And just like what we did for 3π, we can also show that sin−1(21) = 6π.
I think instead of using the word "smallest positive value" , you should use : the smallest solution in the Principal Interval .
Did you know that cos−1cosx=x∀x∈[0,π] ?
I have provided you with the graph of y=cos−1cosx
The honest truth is YOU CAN'T. First of all, arccos is not an algebraic function, and if you want to make it algebraic you will have to turn it into an infinite power series. The issue then becomes how to determine that arccos(21) is EXACTLY 3π (key word being EXACT, because in algebra we don't have any notion of "convergence", whatever that means).