trigonometry: given 0<x<360, prove 3(sin x + cos x )= 2cos

[cai cen ho]

given 0<x<360, prove that 3(sin x + cos x )= 2cos for answer nearest to 0.1

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[Deleted member 4993]

given 0<x<360, prove that 3(sin x + cos x )= 2cos for answer nearest to 0.1

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[stapel]

given 0<x<360, prove that 3(sin x + cos x )= 2cos for answer nearest to 0.1

Since the domain for the variable x is "from zero to three hundred sixty", I'll guess that you're supposed to be working in degrees, rather than radians. I will guess that the left-hand side of the equation, in function notation, is meant to be as follows:

. . . . .\(\displaystyle 3\, \big(\sin(x)\, +\, \cos(x)\big)\)

However, I cannot guess what might possibly be the argument for the cosine on the right-hand side.

Please reply with the missing information. When you reply, please include a clear statement of your efforts so far, so we can see where you're getting stuck. Thank you!