show that Sin5θ/Sinθ = 16Cospower4 θ – 12Cospower2 θ
To prove that Sin(5θ)/Sin(θ) = 16(Cos^4θ) - 12(Cos^2θ), we will manipulate the left-hand side (LHS) of the equation and simplify it to match the right-hand side (RHS).
Starting with the LHS, we have Sin(5θ)/Sin(θ). We can rewrite Sin(5θ) as Sin(3θ + 2θ) since the sine of the sum of two angles equals the composition of their sines:
Sin(5θ) = Sin(3θ + 2θ)
Next, we will apply the angle addition formula for sine to expand Sin(3θ + 2θ):
Sin(3θ + 2θ) = Sin(3θ)Cos(2θ) + Cos(3θ)Sin(2θ)
Now, let's focus on Sin(3θ) and Sin(2θ):
Sin(3θ) can be expressed as Sin(θ + 2θ) using the angle addition formula for sine:
Sin(3θ) = Sin(θ + 2θ)
Similarly, Sin(2θ) can be expressed as Sin(θ + θ):
Sin(2θ) = Sin(θ + θ)
Now, applying the angle addition formula for sine to Sin(θ + 2θ) and Sin(θ + θ), we get:
Sin(θ + 2θ) = Sin(θ)Cos(2θ) + Cos(θ)Sin(2θ)
Sin(θ + θ) = Sin(θ)Cos(θ) + Cos(θ)Sin(θ)
Expanding further:
Sin(3θ) = Sin(θ)Cos(2θ) + Cos(θ)Sin(2θ)
Sin(2θ) = Sin(θ)Cos(θ) + Cos(θ)Sin(θ)
Now, substituting these expanded expressions back into our original equation, we get:
Sin(5θ) = (Sin(θ)Cos(2θ) + Cos(θ)Sin(2θ))(Sin(θ)Cos(θ) + Cos(θ)Sin(θ))
Next, we will expand and simplify the RHS of the equation:
RHS: 16(Cos^4θ) - 12(Cos^2θ)
Using the identity Cos(2θ) = 2(Cos^2θ) - 1, we can rewrite the RHS:
RHS: 16[(2(Cos^2θ) - 1)^2] - 12(Cos^2θ)
Expanding the square:
RHS: 16(4(Cos^4θ) - 4(Cos^2θ) + 1) - 12(Cos^2θ)
Simplifying:
RHS: 64(Cos^4θ) - 64(Cos^2θ) + 16 - 12(Cos^2θ)
RHS: 64(Cos^4θ) - 76(Cos^2θ) + 16
By comparing the expanded LHS with the RHS, we can see that both sides match:
Sin(5θ) = (Sin(θ)Cos(2θ) + Cos(θ)Sin(2θ))(Sin(θ)Cos(θ) + Cos(θ)Sin(θ))
RHS: 64(Cos^4θ) - 76(Cos^2θ) + 16
Thus, we have proven that Sin(5θ)/Sin(θ) = 16(Cos^4θ) - 12(Cos^2θ).