Question
2) We set f(x)=arecos (2x*-1)-2 arccos (x) for all xe|0;1] • a) Calculate f(x) on- [0;1[ b) Deduce that arccos (2x2-1)=2arecos (x) for all x€[0:1] .
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Answer to a math question 2) We set f(x)=arecos (2x*-1)-2 arccos (x) for all xe|0;1] • a) Calculate f(x) on- [0;1[ b) Deduce that arccos (2x2-1)=2arecos (x) for all x€[0:1] .
Esmeralda
4.795 Answers
The function in question is:
f(x) = \arccos(2x^2 - 1) - 2\arccos(x)
Given the trigonometric identity:
\arccos(2x^2 - 1) = 2\arccos(x)
We know that forx \in [0,1]
\arccos(2x^2 - 1) = 2\arccos(x)
Therefore, forx \in [0,1]
f(x) = \arccos(2x^2 - 1) - 2\arccos(x) = 2\arccos(x) - 2\arccos(x) = 0
B.Therefore, the functionf(x) = \arccos(2x^2 - 1) - 2\arccos(x) x [0,1]
\boxed{f(x) = 0}
Given the trigonometric identity:
We know that for
Therefore, for
B.Therefore, the function
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