# Prove $\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$ if $A+B+C=180$ degrees

I most humbly beseech help for this question.

If $A+B+C=180$ degrees, then prove $$\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$$ I am not sure what trig identity I should use to begin this problem.

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try substitution C = 180-A-B to get formulae for $\cos(C)$ and $\sin^2(C)$ in terms of trig functions of $A$ and $B$ –  gt6989b Jul 31 '12 at 15:20
thanks for your post I will study them and use them to solve the problem. –  Fernando Martinez Jul 31 '12 at 15:30
Fernando Martinez appreciates all the post and the work that went into answering his question. –  Fernando Martinez Aug 1 '12 at 15:34

Let's take the Right Hand Side. $$2 \sin B \cos C = \sin(B+C) + \sin(B-C) = \sin A + \sin(B-C)$$ Therefore, $2 \sin A \sin B \cos C = \sin^2 A + \sin A \sin(B-C)$ Now, $$2 \sin A \sin (B-C) = \cos (A - B +C) - \cos (A + B -C) = \cos 2C - \cos 2B = 2 \sin^2 B - 2 \sin ^2 C$$ Cancelling the $2$ gives us therefore, that $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$.

To finish, we already established $2 \sin A \sin B \cos C = \sin^2 A = \sin A \sin (B-C)$. Now, since $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$, therefore, $2 \sin A \sin B \cos C = \sin^2 A + \sin^2 B - \sin^2 C$.

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Hmm which two would I cancel out. –  Fernando Martinez Jul 31 '12 at 15:45
No, I meant to cancel out the $2$ in $2 \sin A \sin (B-C) = 2 ( \sin^2 B - \sin^2 C)$, giving us $\sin A \sin (B-C) = \sin^2 B - \sin^2 C$ –  Rijul Saini Jul 31 '12 at 16:02
I see my now so technically would I have now sin^2B-sin^2C=cos(a-b+c)-cos(a+b-c)=cos2c-cos2b=2sin^2B-2sin^2C –  Fernando Martinez Jul 31 '12 at 16:17
What would I do next I am not sure what to do..... –  Fernando Martinez Jul 31 '12 at 16:20
I edited my post. –  Rijul Saini Jul 31 '12 at 17:10

One can deduce this identity from the conjunction of the law of sines and the law of cosines.

The law of sines says that for the three angles $A$, $B$, $C$ of a triangle, with opposite sides $a$, $b$, $c$, we have $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d.$$ The last equality merely defines $d$, and one can omit it and still have a statement of the law of sines. The common value $d$ is actually the diameter of the circumscribed circle.

If things are scaled so that $d=1$, then we have

$$a=\sin A,\quad b=\sin B,\quad c=\sin C\tag{1}$$

The law of cosines says $$a^2+b^2-2ab\cos C = c^2.\tag{2}$$ Substitute the expressions in $(1)$ into the appropriate places in $(2)$ and you get $$\sin^2 A+\sin^2 B - 2\sin A\sin B\cos C = \sin^2 C$$ and there's your identity.

This still leaves the problem of how to prove the law of sines and the law of cosines. And if you want to use your identity to prove the law of cosines, then this reasoning would be circular. But if you can take the two laws to be established already, then this does it.

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I don't get how the half diameter makes $a = \sin(A)$?i got $\sin(A) = a/.5$; $.5\sin(A) = a$. And i thought $\sin(A) = a$ when the radius was 1? –  Kat Dec 15 '13 at 21:50
The law of sines says $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$. Try it in case $A$ is a right angle, and see that the opposite side is actually the diameter. So $a=1$. I'll write a more leisurely answer if you post that as a separate question. –  Michael Hardy Dec 15 '13 at 21:56

This may not be the shortest way, but is pretty systematic and doesn't involve any real trickery. First, we eliminate the $C$ angles from the equation, using that $C = 180 - A - B$. We write $\sin(C) = \sin(180 - A - B) = \sin(A + B)$, and $\cos(C) = \cos(180 - A - B) = -\cos(A + B)$ and what you need to prove is $$\sin^2A + \sin^2B - \sin^2(A + B) = -2\sin A\sin B \cos(A+B)$$ Inserting the sine and cosine addition formulas in this, what you need to prove becomes $$\sin^2A + \sin^2B - (\sin A\cos B + \cos A \sin B)^2 = -2\sin A \sin B(\cos A \cos B - \sin A \sin B)$$ Writing this out, it becomes $$\sin^2A + \sin^2B - \sin^2 A\cos^2 B - 2\sin A \cos A \sin B \cos B -\cos^2 A \sin^2B$$ $$= -2\sin A\sin B \cos A \cos B + 2\sin^2 A \sin^2 B$$ Cancelling terms, your goal is to prove $$\sin^2A + \sin^2B - \sin^2 A\cos^2 B-\cos^2 A \sin^2B = 2\sin^2 A \sin^2 B$$ Equivalently, $$\sin^2A(1 - \cos^2B) + \sin^2B(1 - \cos^2A) = 2\sin^2 A \sin^2 B$$ This last equation is true since $1 - \cos^2 = \sin^2$. So going backwards in the above steps, one obtains the original equation $\sin^2A + \sin^2B - \sin^2C = 2\sin A \sin B \sin C$ as desired.

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$\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin(B+C)\sin(B-C)$ either using the identity $\sin^2B-\sin^2C=\sin(B+C)\sin(B-C)$

or $\sin^2B-\sin^2C=\frac{1}{2}(2\sin^2B-2\sin^2C)=\frac{1}{2}(1-\cos2B-(1-\cos2C))=\frac{1}{2}(\cos2C-\cos2B)=-\frac{1}{2}2\sin(B+C)\sin(C-B)=\sin(B+C)\sin(B-C)$
as $\sin(-x)=-\sin(x)$

Now, $\sin(B+C)=\sin(180^\circ-A)=\sin{A}$

So, $\sin^2A+\sin^2B-\sin^2C=\sin^2A+\sin{A} \sin(B-C)$

$=\sin{A}(\sin{A}+\sin(B-C))$

$=\sin{A}(\sin(B+C)+\sin(B-C))$ replacing $\sin{A}$ with $\sin(B+C)$

$=\sin{A}(2\sin{B}\cos{C})$

$=2\sin{A}\sin{B}\cos{C}$

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Fernando especially likes this response. –  Fernando Martinez Aug 1 '12 at 16:39
@FernandoMartinez, thanks. Trigonometry is all about formulae & identities. –  lab bhattacharjee Aug 1 '12 at 17:55
@FernandoMartinez, can have a look into math.stackexchange.com/questions/176892/… –  lab bhattacharjee Aug 2 '12 at 16:06

Use:

• $\sin(180 - \alpha) = \sin(\alpha)$, $\cos(180 - \alpha) = -\cos(\alpha)$
• $\sin(\alpha+\beta) = \sin(\alpha) \cos(\beta) + \sin(\beta) \cos(\alpha)$
• $\sin^2(\alpha) + \cos^2(\alpha)=1$

Start with solving for $C$, then use identities from the first bullet. Then expand $\sin(A+B)$ and $\cos(A+B)$ using the identity from the second bullet, and its companion for $\cos(\alpha+\beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha)\sin(\beta)$.

Then simplify the algebraic expressions and use the identity from the third bullet.

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Sasha would I start on the right or left side sorry if its a dumb question.... –  Fernando Martinez Jul 31 '12 at 15:36