# 1) Expand: 2) For which value is f(x) undefined: cos(x)÷(1-cos(2x)) 1, 1/2, π, π/2Please and thanks!

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1) `(2sqrt(2)-pi)^2`

You can expand this using binomal expansion.

= `((2sqrt(2))^2-2(2sqrt(2))pi+pi^2)`

= `(8-4sqrt(2)pi+pi^2)`

Therefore,

`(2sqrt(2)-pi)^2 = (8-4sqrt(2)pi+pi^2)`

2) `f(x) = cos(x)/(1-cos(2x))`

f(x) is undefined when `1-cos(2x)= 0`

`cos(2x) = 1`

Therefore `2x = 0`

The general solution for 2x is given by,

`2x = 2npi+- 0` n is any integer.

`x = npi`

**Therefore, f(x) is undefined only at** `x = pi.`

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#1 is supposed to say:

Expand: (2√2 -π)^2