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Trigonometric function symmetry If the image of y = sin (x + a) + radical three cos (x-a) is symmetric about the Y axis, then the value of a is Why is it wrong to calculate with x = 0, y = plus or minus 2? The answer is a = k π - π / 6
This is a difficult problem. Your mistake is completely expected. People who don't know it think that the topic is wrong. The topic is not only correct, but also imaginative
y=sinx*cosa+cosxsina
RELATED INFORMATIONS
- 1. How to judge whether the domain is symmetric about the origin? For example: for the function f (x) = 3tan (1 / 2 x - π / 3), when discussing the parity of F (x), why is the domain not symmetric about the origin?
- 2. F (x) = 2x & # 178; + 4, X ∈ [- 2,2]
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- 6. Judge the parity of the following functions (1) f (x) = | 1 / 2x-3 | + | 1 / 2x + 3|
- 7. Judging the parity of function f (x) - f (- x)
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- 11. Trigonometric function symmetry If f (x) = sin (π / 4 x + π / 6), G (x) is symmetric with respect to x = 1, then G (x) is obtained There is also an analytic expression of y = 3sin (π / 2x + π / 6) with respect to x = 2 π symmetry
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- 13. Given that the function f (x) has: F (x1x2) = f (x1) + F (x2) for any x1, X2 on R, it is proved that f (x) is an even function
- 14. Let y = f (x) (x belongs to R) satisfy f (x1) + F (x2) = f (x1 * x2) for any real number x1, X2, and prove that (1) f (1) = f (- 1) = 0 (2) f (x) is even function
- 15. Function f (x), X belongs to R, if for any real number x1, X2 has f (x1 + x2) + F (x1-x2) = 2F (x1) f (x2), prove that f (x) is even function
- 16. Let f (x) (x ∈ R) satisfy f (x1 * x2) = f (x1) + F (x2) for any real number x1, X2, and prove that f (x) is an even function
- 17. It is known that the function f (x) is an even function on the set of real numbers R, and is a decreasing function on (0, + ∞). If x 10, then there is A.f（-x1）>f(-x2) B.f(-x1)=f(-x2) c.f(-x1)
- 18. Given the function y = F & nbsp; (x) (x ∈ R, X ≠ 0), for any non-zero real number x1, X2, f (x1x2) = f (x1) + F (x2), try to judge the parity of F (x)
- 19. The function FX = log (1 / 2) (x + 1) / (x-1) is known to judge the parity. It is proved that FX is an increasing function at (1, + ∞)
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