• MHB
• Monoxdifly
In summary, if $\alpha+\beta+\gamma=180°$, then it can be proven that $2\sin\alpha+2\sin\beta+2\sin\gamma=4\sin\alpha\sin\beta\sin\gamma$. This is done by using the trigonometric identity $\sin(\beta+\gamma)=\sin(180°-\alpha)=\sin\alpha$ and simplifying the equation to show that it is an identity.
Monoxdifly
MHB
If $$\displaystyle \alpha+\beta+\gamma=180°$$, prove that $$\displaystyle 2sin\alpha+2sin\beta+2sin\gamma=4sin\alpha sin\beta sin\gamma$$!
All I knew is that $$\displaystyle sin(\beta+\gamma)=sin(180°-\alpha)=sin\alpha$$, but I think it doesn't help in this case.

Monoxdifly said:
If $$\displaystyle \alpha+\beta+\gamma=180°$$, prove that $$\displaystyle \color{red}2sin\alpha+2sin\beta+2sin\gamma=4sin\alpha sin\beta sin\gamma$$!
All I knew is that $$\displaystyle sin(\beta+\gamma)=sin(180°-\alpha)=sin\alpha$$, but I think it doesn't help in this case.

check that equation again, because it's not an identity

Well, this was a question asked by a student I am tutoring. He did say that he tried all angles to be 60° and the equation didn't match, so I told him that it means that it couldn't be proved.

If $x+y+z=\pi$ then

$$\displaystyle \sin2x+\sin2y+\sin2z=4\sin x\sin y\sin z$$

\displaystyle \begin{align*}2(\sin2x+\sin2y+\sin2z)&=\sin x\cos(y-z)+\sin y\cos(x-z)+\sin z\cos(x-y) \\ &=6\sin x\sin y\sin z+\sin x\cos y\cos z+\sin y\cos x\cos z+\sin z\cos x\cos y \\ &=6\sin x\sin y\sin z+\sin x\cos x+\sin y\cos y+\sin z\cos z \\ \frac32(\sin2x+\sin2y+\sin2z)&=6\sin x\sin y\sin z \\ \sin2x+\sin2y+\sin2z&=4\sin x\sin y\sin z\end{align*}

## 1. How do you prove the equation 2sinα+2sinβ+2sinγ=4sinαsinβsinγ in trigonometry?

To prove this equation, we can use the trigonometric identity sin(A+B) = sinAcosB + cosAsinB. By applying this identity, we can rewrite the left side of the equation as 2(sinαcosβ + cosαsinβ) + 2sinγ. Then, using the identity again, we can further simplify to 2sin(α+β) + 2sinγ. Finally, we can use the identity sin(A+B+C) = sinAcosBcosC + cosAsinBcosC + cosAcosBsinC + sinAsinBsinC to rewrite the right side of the equation as 4sinαsinβsinγ. This shows that both sides are equal and proves the equation.

## 2. Can you use the Pythagorean identity to prove 2sinα+2sinβ+2sinγ=4sinαsinβsinγ?

No, the Pythagorean identity (sin^2x + cos^2x = 1) is used to simplify trigonometric expressions, but it cannot be used to prove equations like this one. To prove this equation, we need to use the trigonometric identities for sum and product of angles.

## 3. What are the steps to prove 2sinα+2sinβ+2sinγ=4sinαsinβsinγ using the double angle formula?

The double angle formula for sine is sin2x = 2sinxcosx. To prove this equation, we can use this formula to rewrite the left side as 2sinαcosα + 2sinβcosβ + 2sinγcosγ. Then, using the identity sin(A+B) = sinAcosB + cosAsinB, we can rewrite each term as sin(2α) + sin(2β) + sin(2γ). Finally, we can use the identity sin(A+B+C) = sinAcosBcosC + cosAsinBcosC + cosAcosBsinC + sinAsinBsinC to rewrite the right side of the equation as 4sinαsinβsinγ. This shows that both sides are equal and proves the equation.

## 4. Is there a geometric proof for 2sinα+2sinβ+2sinγ=4sinαsinβsinγ?

Yes, there is a geometric proof for this equation. It involves constructing a triangle with angles α, β, and γ and using the Law of Sines to show that the left side of the equation is equal to the right side. This proof can be found in many trigonometry textbooks or online resources.

## 5. Can you prove 2sinα+2sinβ+2sinγ=4sinαsinβsinγ without using trigonometric identities?

No, this equation involves trigonometric functions and can only be proven using trigonometric identities. Without using these identities, it is not possible to simplify the equation and show that both sides are equal.

Replies
2
Views
1K
Replies
4
Views
935
Replies
2
Views
2K
Replies
4
Views
1K
Replies
5
Views
2K
Replies
1
Views
299
Replies
3
Views
840
Replies
8
Views
496
Replies
9
Views
2K
Replies
2
Views
2K