# Calculate determinant by expansion and problem of substitution [(1,cosa,cos2a)][(1,cosb,cos2b)][(1,cosc,cos2c)]

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The determinant has three rows and columns and it could be evaluated using Sarrus' rule.

To make the evaluation easier you should use the properties of determinants.

`[[1,cos a , cos 2a],[1,cosb , cos 2b],[1,cosc , cos 2c]]`

Replacing each element of the third column by the formula`cos 2 alpha = 2cos^2 alpha -1 ` yields:

`[[1,cos a ,2cos^2 a - 1],[1,cosb , 2cos^2b - 1],[1,cosc , 2cos^2c - 1]]`

Adding the first column to the third column yields:

`[[1,cos a ,2cos^2 a],[1,cosb , 2cos^2b],[1,cosc , 2cos^2c]]`

Notice the common factor 2:

`[[1,cos a ,2cos^2 a],[1,cosb , 2cos^2b],[1,cosc , 2cos^2c]] = 2*[[1,cos a ,cos^2 a],[1,cosb , cos^2b],[1,cosc , cos^2c]]`

The new determinant is a Vandermonde determinant.

`[[1,cos a ,cos^2 a],[1,cosb , cos^2b],[1,cosc , cos^2c]] = (cos b - cos a)(cos c - cos a)(cos c - cos b)`

**Evaluating the determinant yields `Delta = 2(cos b - cos a)(cos c - cos a)(cos c - cos b).` **