Question

Please help, another math problem! This is all the information it gives me!
1+cos(6x)=

a. 2cos^2(3x)
b. 2sin^2(3x)
c. 3sin(2x)
d. 3cos(2x)

Answer 1

$1 + cos(6x) = 1 + cos(3x + 3x)$
$= 1 + cos3x cos3x - sin3xsin3x$
$= 1 + cos^{2}(3x) - sin^{2}(3x)$
$= 1 + cos^{2}(3x) - (1 - cos^{2}(3x))$
$= 2cos^{2}(3x)$

Answer should be (A), to be honest.

Answer 2

An expression is defined as a set of numbers, variables, and mathematical operations. The value of the given algebraic expression 1 + cos(6x) can be written as 2cos²(3x). The correct option is A.

What is an Expression?

In mathematics, an expression is defined as a set of numbers, variables, and mathematical operations formed according to rules dependent on the context.

Given the algebraic expression, 1+cos(6x), the value of the expression can be calculated using the algebraic identity cos(2x)= 2cos²(x)-1. Therefore, the value of the expression can be written as,

1 + cos(6x)

= 1 + 2cos²(3x) - 1

= 2cos²(3x)

Hence, the value of the given algebraic expression 1 + cos(6x) can be written as 2cos²(3x).

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