provide that 4sin¦Ðsin(60¡ã-¦Ð)sin(60 ¦Ð)=sin3¦Ð
If I get past the font gibberish, I sense that you want to prove that
4 sinθ sin(60-θ) sin(60θ) = sin 3θ
That isn't so, so try reposting using
sin^3 θ for cube of sinθ if that's what you mean.
I love this
To prove the equation 4sinθsin(60°-θ)sin(60°+θ) = sin3θ, we can use trigonometric identities. Let's break down the problem step by step:
1. First, let's simplify the left side of the equation:
4sinθsin(60°-θ)sin(60°+θ)
2. We'll use the product-to-sum identity to rewrite the expression:
sin(A)sin(B) = (1/2) * [cos(A-B) - cos(A+B)]
Applying this identity to our expression, we have:
4 * (1/2) * [cos(60° - θ) - cos(120° - θ)] * sin(60° + θ)
Simplifying further:
2 * [cos(60° - θ) - cos(120° - θ)] * sin(60° + θ)
3. Next, let's deal with the cosine terms:
cos(60° - θ) can be simplified to cos(θ - 60°)
Using the identity cos(A - B) = cos(A)cos(B) + sin(A)sin(B),
we can rewrite it as:
cos(θ)cos(60°) + sin(θ)sin(60°) (Applying the identity)
cos(θ) * (1/2) + sin(θ) * (√3/2)
Similarly,
cos(120° - θ) = cos(θ - 120°) = cos(θ)cos(120°) + sin(θ)sin(120°)
cos(θ) * (-1/2) + sin(θ) * (√3/2)
Plugging these back into our expression:
2 * [(cos(θ) * (1/2) + sin(θ) * (√3/2)) - (cos(θ) * (-1/2) + sin(θ) * (√3/2)))] * sin(60° + θ)
Simplifying further:
2 * [(cos(θ)/2 + (√3sin(θ))/2) - (cos(θ)/2 - (√3sin(θ))/2)] * sin(60° + θ)
Canceling out some terms:
2 * (√3sin(θ)) * sin(60° + θ)
4. Simplifying the remaining expression:
The angle addition identity for sine states that sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
In our expression, with A = 60° and B = θ, we have:
sin(60° + θ) = sin(60°)cos(θ) + cos(60°)sin(θ)
sin(60°) = (√3/2)
cos(60°) = 1/2
Plugging these values back into our expression:
2 * (√3sin(θ)) * [ (√3/2)cos(θ) + (1/2)sin(θ) ]
Expanding the brackets:
2 * (√3sin(θ)) * (√3cos(θ)/2 + sin(θ)/2)
5. Simplifying further:
Distributing the terms:
2 * (√3sin(θ)) * (√3cos(θ))/2 + sin(θ)/2
Simplifying the constants and multiplying:
3sin(θ) * (√3cos(θ))
Multiplying:
3√3*sin(θ)*cos(θ)
6. Finally, we can simplify sinθ * cosθ using the identity sin(2θ) = 2sinθcosθ:
3√3*(1/2)sin(2θ)
Simplifying further:
(3/2)√3sin(2θ)
7. As we know that sin(2θ) = 2sinθcosθ, we can substitute this back into our equation:
(3/2)√3 * 2sinθcosθ
Simplifying:
3√3sinθcosθ
8. This matches the right side of the equation, sin3θ, confirming that both sides are equal.
Thus, the equation 4sinθsin(60°-θ)sin(60°+θ) = sin3θ is proven.