Possible Duplicate: Prove that $sin(2A)+sin(2B)+sin(2C)=4sin(A)sin(B)sin(C)$ when $A,B,C$ are angles of a triangle Prove trigonometry identity?

If $A$, $B$, and $C$ are to lớn be taken as the angles of a triangle, then I beg someone lớn help me the proof of$$sin A + sin B + sin C = 4 cos fracA2 cos fracB2 cos fracC2.$$Thanks!

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I usually find these things easiest to bởi with complex exponentials, rather than remembering a zoo of formulas for trigonometric functions. With $C=pi-A-B$, we have

$$eginaligndef r#1#2gamize.vnrm e^gamize.vnrm i#1#2gamize.vnrm e^-gamize.vnrm i#1defco#1 r#1+defsi#1 r#1-defcop#1left(co#1 ight)defsip#1left(si#1 ight)&4cosfrac A2cosfrac B2cosfrac C2\=½copA/2copB/2copC/2\=½copA/2copB/2cop(pi-A-B)/2\=½gamize.vnrm icopA/2copB/2sip(-A-B)/2\=½gamize.vnrm ileft(-sip A-sip B-sip(A+B) ight)\=½gamize.vnrm ileft(sip A+sip B+sip(pi-A-B) ight)\=½gamize.vnrm ileft(sip A+sip B+sip C ight)\=&sin A+sin B+sin C;.endalign$$

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answered Aug 9, 2012 at 23:14


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$egingroup$ +1 for giving a general strategy for doing this type of thing, rather than just the straight proof. $endgroup$
Aug 10, 2012 at 0:47
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How about a proof with a geometric flavor?

Let $a$, $b$, $c$ be the sides that oppose angles $A$, $B$, $C$, respectively. By the Law of Sines, $$fracasin A = fracbsin B = fraccsin C = d$$where $d$ is the circumdiameter of the triangle. If we conveniently scale the triangle so that $d=1$, then we can say simply that $$a = sin A qquad b = sin B qquad c = sin C$$This is a common simplification technique, as it nicely blurs the distinction between edges and angles, giving us things lượt thích this wonderfully-symmetric area formula:$$gamize.vnrmArea ; = frac12a b c qquad left(;=frac12absin C =frac12acsin B = frac12bcsin A ight)$$

(When you have this mindset, you can"t look at the expression "$sin A + sin B + sin C$" và not think, "That"s perimeter!" ... Và then you find yourself pursuing proof approaches like this one.) In what follows, I"ll continue to lớn use "$a$", "$b$", "$c$", because they"re more compact than "$sin A$", etc, but you should read them as "$sin A$", etc.

By the Law of Cosines,

$$cosC = fraca^2+b^2-c^22ab$$

By the half-angle formula for cosines,

$$cos^2fracC2 = frac1+cosC2=fraca^2+2ab+b^2-c^24ab=frac(a+b)^2-c^24ab=frac(a+b+c)(a+b-c)4ab$$

Likewise for $cos(A/2)$ & $cos(B/2)$, so that$$cos^2fracA2 cos^2fracB2 cos^2fracC2=frac(a+b+c)^24a^2b^2c^2cdot frac(a+b+c)(-a+b+c)(a-b+c)(a+b-c)16$$The conveniently-separated second factor just happens to be Heron"s formula for the square of the area of the triangle; re-writing the area in wonderfully-symmetric khung gives ...$$cos^2fracA2 cos^2fracB2 cos^2fracC2=frac(a+b+c)^24a^2b^2c^2cdot left(frac12abc ight)^2 = frac116left(a+b+c ight)^2$$

We can now clear the fraction, expand the symbols "$a$", "$b$", "$c$" as the sines they represent, và take square roots (secure in the knowledge that none of the trig values is negative), so that$$4 cos fracA2 cosfracB2 cosfracC2 = sin A + sin B + sin C$$as desired.