prove that
(2cosx+1)(2cosx-1)= 1 +2cos2x
Using
cos(2x)=cos²(x)-sin²(x)
and cos²(x)=1-sin²(x)
expand the right-hand-side to
4cos²(x)-1
which easily factorizes to the left-hand-side, namely:
(2cosx+1)(2cosx-1)
How did you find 4cos^2-1 when expanding the right side
To prove the given equation, we'll expand the left-hand side and simplify it to see if it equals the right-hand side.
Starting with the left-hand side: (2cosx + 1)(2cosx - 1)
Using the FOIL method (first, outer, inner, last), we multiply each term:
= 2cosx * 2cosx + 2cosx * (-1) + 1 * 2cosx + 1 * (-1)
= 4cos^2(x) - 2cosx + 2cosx - 1
The middle terms (-2cosx + 2cosx) cancel each other out:
= 4cos^2(x) - 1
Now, let's simplify the right-hand side:
1 + 2cos(2x)
We know that the double-angle formula for cosine is cos(2x) = 2cos^2(x) - 1.
By substituting this formula into the equation, we get:
1 + 2(2cos^2(x) - 1)
= 1 + 4cos^2(x) - 2
Now, combining like terms:
= 4cos^2(x) - 1
We observe that the right-hand side equals the simplified expression we obtained from the left-hand side.
Therefore, we have proven that (2cosx + 1)(2cosx - 1) = 1 + 2cos(2x).
To prove the given equation: (2cosx+1)(2cosx-1) = 1 + 2cos2x.
Let's start by expanding the left side of the equation:
(2cosx + 1)(2cosx - 1) = 4cos^2(x) - 2cosx - 2cosx + 1 = 4cos^2(x) - 4cosx + 1
Now, let's simplify the right side of the equation:
1 + 2cos2x
To simplify this further, we'll use the double-angle formula for cosine:
cos2x = 2cos^2(x) - 1
Substituting this into the equation:
1 + 2cos2x = 1 + 2(2cos^2(x) - 1) = 1 + 4cos^2(x) - 2 = 4cos^2(x) - 1
Now, we see that the left side equals 4cos^2(x) - 4cosx + 1, and the right side equals 4cos^2(x) - 1. Since both sides are equal, we have proved that:
(2cosx + 1)(2cosx - 1) = 1 + 2cos2x.