# Write 1+cosx as a product .Write 1+cosx as a product .

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You may also replace `cos 2pi` for 1, such that:

`1 + cos x = cos 2pi + cos x`

Converting the summation into a product, using the formula `cos alpha + cos beta = 2 cos((alpha + beta)/2)*cos((alpha - beta)/2)` , yields:

`cos 2pi + cos x = 2 cos((2pi + x)/2)*cos((2pi - x)/2)`

`cos 2pi + cos x = 2 cos(pi + x/2)*cos(pi - x/2)`

Since `cos(pi + x/2) = cos(pi - x/2) = -cos (x/2)` yields:

`cos 2pi + cos x = 2*(-cos (x/2)*(-cos (x/2))`

`cos 2pi + cos x = 2* cos^2(x/2)`

You also may use the half angle identity, to convert the summation `1 + cos x` into a product, such that:

`cos(x/2) = sqrt((1 + cos x)/2) => 2cos^2(x/2) = 1 + cos x`

**Hence, converting the summation `1 + cos x` into a product, yields **`2cos^2(x/2) = 1 + cos x.`

To transform the sum into a product, we'll have to have in the given expression, 2 like trigonometric functions.

Since we have already the term cos x, we'll write the value 1 as cos 0.

We'll re-write the expression:

cos 0 + cos x

To transform the sum into a product we'll apply the formula:

cos a + cos b = 2cos[(a+b)/2]cos[(a-b)/2]

We'll put a = 0 and b = x:

cos 0 + cos x = 2cos[(0+x)/2]cos[(0-x)/2]

cos 0 + cos x = 2cos(x/2)*cos(-x/2)

Since the cosine function is even, we'll get:

1 + cos x = 22cos(x/2)*cos(x/2)

1 + cos x = 2[cos(x/2)]^2