1. ## Trig Triangle Proof

Hello Everyone,

I have another question again...

In any acute-angle triangle ABC, prove that:
b^2sin2C + c^2sin2B = 2bcsinA

Does it have to do with sine law and cosine law? I can't figure it out.

2. Originally Posted by KelvinScale
Hello Everyone,

I have another question again...

In any acute-angle triangle ABC, prove that:
b^2sin2C + c^2sin2B = 2bcsinA

Does it have to do with sine law and cosine law? I can't figure it out.
$b^2\sin{2C} + c^2\sin{2B} = 2b^2\sin{C}\cos{C} + 2c^2\sin{B}\cos{B}$

Now from the cosine rule:

$c^2 = a^2 + b^2 - 2ab\cos{C}$

$2ab\cos{C} = a^2 + b^2 - c^2$

$\cos{C} = \frac{a^2 + b^2 - c^2}{2ab}$.

Similarly $\cos{B} = \frac{a^2 + c^2 - b^2}{2ac}$.

So $2b^2\sin{C}\cos{C} + 2c^2\sin{B}\cos{B} = 2b^2\sin{C}\left(\frac{a^2 + b^2 - c^2}{2ab}\right) + 2c^2\sin{B}\left(\frac{a^2 + c^2 - b^2}{2ac}\right)$

$= \frac{b(a^2 + b^2 - c^2)\sin{C}}{a} + \frac{c(a^2 + c^2 - b^2)\sin{B}}{a}$

$= \frac{b(a^2 + b^2 - c^2)\sin{C} + c(a^2 + c^2 - b^2)\sin{B}}{a}$

We also know from the sine rule:

$\frac{\sin{C}}{c} = \frac{\sin{A}}{a}$

so $\sin{C} = \frac{c\sin{A}}{a}$.

Similarly $\sin{B} = \frac{b\sin{A}}{a}$.

So $\frac{b(a^2 + b^2 - c^2)\sin{C} + c(a^2 + c^2 - b^2)\sin{B}}{a}$

$= \frac{b(a^2 + b^2 - c^2)\left(\frac{c\sin{A}}{a}\right) + c(a^2 + c^2 - b^2)\left(\frac{b\sin{A}}{a}\right)}{a}$

$= \frac{bc\sin{A}(a^2 + b^2 - c^2)}{a^2} + \frac{bc\sin{A}(a^2 + c^2 - b^2)}{a^2}$

$= \frac{a^2bc\sin{A}}{a^2} + \frac{b^3c\sin{A}}{a^2} - \frac{bc^3\sin{A}}{a^2} + \frac{a^2bc\sin{A}}{a^2} + \frac{bc^3\sin{A}}{a^2} - \frac{b^3c\sin{A}}{a^2}$

$= \frac{2a^2bc\sin{A}}{a^2}$

$= 2bc\sin{A}$.

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# b×bsin2C c×csin2B=2bcsinA proof

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