By expressing 3x as (2x + x) prove that
cos3x = 4cos^3x - 3cosx
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To prove this trigonometric identity using the expression 3x as (2x + x), we can start with the double-angle formula for the cosine function, which states:
cos(2θ) = 2cos^2(θ) - 1
Let's now replace θ with x,:
cos(2x) = 2cos^2(x) - 1
Now, we can express 3x as (2x + x):
cos(2(2x + x)) = 2cos^2(2x + x) - 1
Simplifying the left side:
cos(4x + 2x) = 2cos^2(3x) - 1
Using the sum-to-product formula for cosine, we can rewrite the left side:
cos(6x) = 2cos^2(3x) - 1
Now, we can express cos(6x) using cos(3x) and cos(x):
cos(6x) = 2cos^2(3x) - 1
cos(2(3x)) = 2cos^2(3x) - 1
Applying the double-angle formula again:
2cos^2(3x) - 1 = 2(2cos^2(3x) - 1)^2 - 1
We can simplify this further by squaring the expression:
2cos^2(3x) - 1 = 2(4cos^4(3x) - 4cos^2(3x) + 1) - 1
Distributing the 2:
2cos^2(3x) - 1 = 8cos^4(3x) - 8cos^2(3x) + 2 - 1
Combining like terms:
2cos^2(3x) - 1 = 8cos^4(3x) - 8cos^2(3x) + 1
Rearranging the terms:
8cos^4(3x) - 8cos^2(3x) + 2 - 1 = 2cos^2(3x) - 1
Simplifying:
8cos^4(3x) - 10cos^2(3x) + 1 = 0
Now, let's define a new variable t = cos(3x):
8t^4 - 10t^2 + 1 = 0
This is a quadratic equation in t^2. We can solve it by factoring or using the quadratic formula:
(4t^2 - 1)(2t^2 - 1) = 0
The solutions for t^2 are:
t^2 = 1/4 or t^2 = 1/2
Taking the square root of both sides:
t = ±√(1/4) or t = ±√(1/2)
Simplifying:
t = ±1/2 or t = ±1/√2
Substituting back t = cos(3x):
cos(3x) = ±1/2 or cos(3x) = ±1/√2
Hence, we have proved that cos(3x) = 4cos^3(x) - 3cos(x) using the given expression 3x as (2x + x).