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Write 1+cosx as a product .Write 1+cosx as a product .
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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
Posted by giorgiana1976 on June 4, 2011 at 4:23 AM (Answer #2)
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.`
Posted by sciencesolve on April 30, 2013 at 6:04 PM (Answer #3)
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