Respuesta :
Answer:
[tex]B=12x[/tex]
Step-by-step explanation:
Given: [tex]cos^2(6x)-sin^2(6x)=cos(B)[/tex]
Let's simplify the left part of the equation:
[tex]cos(12x)=cos(B)[/tex]
Solve for B:
[tex]B=12x[/tex]
The value of B which for the expression cos^2(6x)-sin^2(6x)=cos(B) is B = 12x + 2πn
What is the expansion of sum for cosine of sum of two angles?
Suppose we've got cosine of angle A and B measurements.
Then, we have:
[tex]\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)[/tex]
If we get A = B, then:
[tex]\cos(2A) = \cos^2(A)- \sin^2(A)[/tex]
Since we're specified the equation [tex]cos^2(6x)-sin^2(6x)=cos(B)[/tex]
Thus, by the above specified conclusion, we get;
[tex]\cos^2(6x)-\sin^2(6x)=\cos(B)\\\\\cos(2 \times 6x) = \cos(B) \\\\\cos(12x) = \cos(B)[/tex]
Since cosine has period of [tex]2\pi[/tex] radians, therfore we get:
[tex]B = 12x + 2\pi n[/tex]
where n is an integer.
Thus, the value of B which for the expression cos^2(6x)-sin^2(6x)=cos(B) is B = 12x + 2πn
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