My name is Kate and I'm a student in high school (10-12).

Please help me verify the identity:
cos²x(sec²x-1)=sin²x

Also I am having trouble withdetermining whether f(x) is odd, even, or neither
f(x)=x³-x

Hi Kate,

What I see here is the trig identity

tan² x + 1 = sec² x

and hence

sec² x - 1 = tan² x

Thus

cos²x(sec² x - 1) = cos²x (tan² x)

Can you complete it now?

In the second problem I would experiment a little. If f(x) is even then for any number x,

f(x) = f(-x)

Choose some numbers x and try.

If x = 1 then f(1) = 0 and f(-1) = 0
If x = 2 then f(2) = 8 - 2 = 6 and f(-2) = -8 -(-2) = -6

Thus f(x) is not even because f(2) is not equal to f(-2).

But f(2) = -f(-2) and the requirement that f(x) be odd is that for all numbers x, f(x) = -f(-x). Is that true, if f(x) = x³ - x is f(x) = -f(-x)? Can you prove it?

Penny

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