# triangle trigonometry

In triangle $ABC$,if $\sin^2 \alpha+ \sin^2 \beta = \sin \gamma$ and $\alpha$ and $\beta$ are acute show that $\gamma= 90$ degree.  I try following: $$\dfrac{1-\cos 2\alpha}{2}+\dfrac{1-\cos 2 \beta}{2}=\sin (\alpha+\beta)$$ $$1-\cos(\alpha+\beta)\cos(\alpha-\beta)=\sin(\alpha+\beta)$$ Now,i want to factorize one side of equation,and on the other i want to have 0 or 1.But i don't know how to do it

• are you sure it is not $\sin^2 \alpha + \sin^2 \beta = \sin^2 \gamma?$ – abel Apr 7 '15 at 20:08
• @abel, nope, it works even with the given condition. Though, if it is the condition you gave, the proof becomes more straightforward. – Prasun Biswas Apr 7 '15 at 20:09
• i am sure that is $\sin \gamma$ – chaos Apr 7 '15 at 20:10

$$1-\sin\gamma=1-[\sin^2\alpha+\sin^2\beta]=\cos^2\alpha-\sin^2\beta=\cos(\alpha+\beta)\cos(\alpha-\beta)$$

Now $\cos(\alpha+\beta)=\cos(\pi-\gamma)=-\cos\gamma$

$$\implies1-\sin\gamma=-\cos\gamma\cos(\alpha-\beta)$$

$$\iff\left(\cos\dfrac\gamma2-\sin\dfrac\gamma2\right)^2+\left(\cos\dfrac\gamma2-\sin\dfrac\gamma2\right)\left(\cos\dfrac\gamma2+\sin\dfrac\gamma2\right)\cos(\alpha-\beta)=0$$

$$\iff\left(\cos\dfrac\gamma2-\sin\dfrac\gamma2\right)\left[\left(\cos\dfrac\gamma2-\sin\dfrac\gamma2\right)+\left(\cos\dfrac\gamma2+\sin\dfrac\gamma2\right)\cos(\alpha-\beta)\right]=0$$

If $\cos\dfrac\gamma2-\sin\dfrac\gamma2=0\iff\tan\dfrac\gamma2=1=\tan\dfrac\pi4\iff\dfrac\gamma2=n\pi+\dfrac\pi4\iff\gamma=2n\pi+\dfrac\pi2$ where $n$ is any integer

As $0<\gamma<\pi,n=0$

Else $\left(\cos\dfrac\gamma2-\sin\dfrac\gamma2\right)+\left(\cos\dfrac\gamma2+\sin\dfrac\gamma2\right)\cos(\alpha-\beta)=0$

$\iff\cos(\alpha-\beta)=\cdots=\tan\left(\dfrac\gamma2-\dfrac\pi4\right)\ \ \ \ (1)$

We need $-1<\tan\left(\dfrac\gamma2-\dfrac\pi4\right)\le1$

Clearly, $(1)$ does not have any unique solution.

Hint:

Use the sine rule which is,

$$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}$$

where $\alpha,\beta,\gamma$ are the angles and the corresponding opposite sides to the angles are $a,b,c$ respectively.

After some grunt work and using the given conditions, you'll get $\sin\gamma = \dfrac{c^2}{a^2+b^2}$

Assume that $\gamma=90^{\circ}$. Then, it is a right triangle with hypotenuse $c$ and you see that the obtained equation holds since both LHS and RHS will yield $1$ (thanks to the Pythagorean theorem for right triangles).

Hence, our assumption is true and we have $\gamma=90^{\circ}$

• Do you need the grunt work in the answer or can you take it home from here? – Prasun Biswas Apr 7 '15 at 20:08
• i can take it,but why you can assume that $\gamma=90$ – chaos Apr 7 '15 at 20:13
• @chaos, Actually, it'd have been a more straight-forward proof if the condition given was the one abel stated. But since it isn't, we need to assume. I can't see any other possible way without assuming that $\gamma=90^{\circ}$ to conclude the proof. – Prasun Biswas Apr 7 '15 at 20:16
• actually,i think that $\sin \gamma=\dfrac{c^2}{a^2+b^2}$ is correct – chaos Apr 7 '15 at 20:18
• @chaos, sorry. it's a typo. I'll fix it. Thanks for pointing it out. :D – Prasun Biswas Apr 7 '15 at 20:19