what does cos(2*arccos(-2/√(5)) equal?
I have the equation cos2x where we substitute arccos(-2/√(5)) for the X. I know it doesn't equal 2 *-2/√(5) So then how do I cancel out the cos? because i need to get rid of all trig functions in my answer.
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Since cos^2A = (1+cos2A)/2 , A= arcos (-2/√(5))
cos 2A= 2cos^2A-1
(cos A)^2 = 4/5
cos 2A= 8/5-1 = 3/5
cos 2X, where X =arccos(-2/sqrt5)
2*cos^2(X) - 1
[cos(arccos(-2/â(5))]^2
those cancel
2*[-2/â(5)]^2-1 = .6
it's the same answer when you plug the entire thing into calculator. I hope this makes sense to you :)
Recall that cos(2x) = cos²(x) - sin²(x). Substituting arccos(-2/â5) for the x, we get:
cos²(arccos(-2/â5)) - sin²(arccos(-2/â5)
= (-2/â5)² - sin²(arccos(-2/â5))
Let z = arccos(-2/â5) => cos(z) = -2/â5
sin²(z) = sin²(arccos(-2/â5)) = (1/â5)² = 1/5
(-2/â5)² - 1/5 = 3/5 = 0.6