How can we prove the identity cos(3x) = 4cos^3(x) – 3cos(x)?

The trigonometric identity we want to prove is cos(3x) = 4cos^3(x) – 3cos(x). This identity relates the cosine of triple angles to the cosine of a single angle.

To demonstrate this identity, we can employ the angle addition formula for cosine. The cosine of the sum of two angles can be expressed as:

cos(A + B) = cos(A)cos(B) – sin(A)sin(B)

Using this formula, we can find cos(3x) by rewriting it as:

cos(3x) = cos(2x + x)

Applying the angle addition formula:

cos(3x) = cos(2x)cos(x) – sin(2x)sin(x)

Next, we need to express cos(2x) and sin(2x) in terms of cos(x). Using the double angle formulas:

cos(2x) = 2cos^2(x) – 1

sin(2x) = 2sin(x)cos(x)

Substituting these into our expression yields:

cos(3x) = (2cos^2(x) – 1)cos(x) – (2sin(x)cos(x))sin(x)

This simplifies to:

cos(3x) = (2cos^3(x) – cos(x)) – 2sin^2(x)cos(x)

Recognizing that sin^2(x) = 1 – cos^2(x), we can substitute this in:

cos(3x) = 2cos^3(x) – cos(x) – 2(1 – cos^2(x))cos(x)

Which simplifies to:

cos(3x) = 2cos^3(x) – cos(x) – 2cos(x) + 2cos^3(x)

Combining like terms, we have:

cos(3x) = 4cos^3(x) – 3cos(x)

This proves the identity cos(3x) = 4cos^3(x) – 3cos(x). Thus, our proof is complete and we have shown that the equation holds true for values of x.

Leave a Comment