Answer:
[tex]\cos(19 \times 2) = \cos(38)[/tex]
Step-by-step explanation:
[tex]\cos^2x - \sin^2x = \cos(2x)[/tex]
Let's prove this.
Using the trigonometric identity of addition,
[tex]\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y)[/tex]
We can prove the previous identity by substituting y = x.
[tex]x = y \to \cos(x + y)\\\\\cos(2x) = \cos(x + x) = \cos(x)\cos(x) - \sin(x)\sin(x)\\\to \cos(2x) = \cos^2(x)-\sin^2(x)[/tex]
Therefore,
[tex]x = 19 \to \cos^2(x) - \sin^2(x)\\\\\cos^2(19) - \sin^2(19) = \cos(2 \times 19) = \cos(38)[/tex]
