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How do you determine if #F(x)= sin x + cos x# is an even or odd function?

Answer 1

Christian Antunez

It is neither.

To be even the function must obey:

#f(-x) = f(x)#

To be odd, the function must obey:

#f(-x) = -f(x)#

In this case,

#F(-x) = sin(-x) + cos(-x) = -sin(x) + cos(x)#

#F(-x) != F(x) or -F(x)# hence it is neither even or odd.

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Answer 2

Serenity Jones

To determine if ( F(x) = \sin(x) + \cos(x) ) is an even or odd function, we evaluate ( F(-x) ) and compare it to ( F(x) ).

For an even function: [ F(-x) = F(x) ]

For an odd function: [ F(-x) = -F(x) ]

Let's evaluate ( F(-x) ) for ( F(x) = \sin(x) + \cos(x) ):

[ F(-x) = \sin(-x) + \cos(-x) ]

Using the properties of sine and cosine functions: [ \sin(-x) = -\sin(x) ] [ \cos(-x) = \cos(x) ]

Substituting these values back into ( F(-x) ): [ F(-x) = -\sin(x) + \cos(x) ]

Comparing ( F(-x) ) to ( F(x) = \sin(x) + \cos(x) ): [ F(-x) \neq F(x) ]

[ F(-x) \neq -F(x) ]

Since neither condition for even nor odd functions is satisfied, ( F(x) = \sin(x) + \cos(x) ) is neither an even nor an odd function.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.