Algebra Precalculus

I want lớn prove sầu this (where each angle may be negative sầu or greater than $180^circ$):

When $A+B+C = 180^circ$ eginequation* an A + ung B + ung C = ã A: ung B: ã C. endequation*

We know thateginequation* an(A+B) = frac an A+ ung B1- ung Achảy Bendequation* và that eginequation* extvà that~A+B = 180^circ-C.endequation*

Therefore $ an(A+B) = - an C.$

From here, I got stuông chồng.

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cảnh báo that$$ hithptquocgia2016.comrmImleft(e^ipi ight)=0 ag1$$Thus, if $a+b+c=pi$,$$eginalign0&= hithptquocgia2016.comrmImleft(e^iae^ibe^ic ight)\&= hithptquocgia2016.comrmImBig(ig(cos(a)+isin(a)ig)ig(cos(b)+isin(b)ig)ig(cos(c)+isin(c)ig)Big)\<4pt>&=sin(a)cos(b)cos(c)+cos(a)sin(b)cos(c)+cos(a)cos(b)sin(c)\&-sin(a)sin(b)sin(c) ag2endalign$$Dividing $(2)$ by $cos(a)cos(b)cos(c)$ yields$$ an(a)+ an(b)+ an(c)= an(a) an(b) an(c) ag3$$


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HINT

$A+B+C = 180$

$A+B = 180 - C$

We"ll apply tangent function:

$ ã (A+B) = chảy (180 - C)$

We"ll consider the identity:

$ an(x+y) = fracchảy x + ã y1-chảy xchảy y$

$frac ã A + chảy B1- ã A an B = fracchảy 180 - ung C1+ ã 180 ung C$

But $ ã 180 = 0$, therefore, we"ll get:

$fracchảy A + chảy 1- ã A an B$ = $frac0 - chảy C1+0$

$fracchảy A + ã B1- ã A ung B = -chảy C$

We"ll multiply by $(1- ung Achảy B)$:

$ ã A + chảy B = -chảy C + an A ung B an C$

Hence

$chảy A + an B+ an C = an A an B ã C$


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answered Aug 27 "13 at 14:22
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ShobhitShobhit
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Here is a geometric proof, for the case that all three angles are acute:

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$QRUV$ are collinear because $B+90^circ+(90^circ-B)=180^circ$.

$STV$ are collinear because $A+B+C=180^circ$, so $angle QSV=angle UTV=C$.

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Similar triangles $ riangle PQRsim riangle TRS$ & $ riangle RTU syên riangle SRQ$ give sầu $displaystylefracQPRQ = fracRTSR = fracTURQ$, and therefore $TU=QP=1$.

Then,$$eginalign& ã A + ã B + ã C = QR+RU+UV = QV \&= QP.. fracQRQP, fracQSQR , fracQVQS = 1 cdot an(A) an(B) an(C) endalign$$

When one of the angles is obtuse, let it (without loss of generality) be $C$. Then a similar diagram can be drawn, except that $V$ is to lớn the left of $Q$, và $UV$, $QV$ count as negative lengths.


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edited Jun 11 "15 at 14:54
answered May 17 "15 at 13:55
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hmakholm left over Monicahmakholm left over Monica
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HINT:

Using $displaystyle an(A+B)=fracchảy A+chảy B1- an A ã B,$

we can prove sầu $$ an(A+B+C)=fracsum_ extcycchảy A-prod ã A1-sum_ extcycchảy Achảy B$$

Now, if $A+B+C=n180^circ,$ where $n$ is any integer we know $ an(n180^circ)=0$


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edited Nov 16 "13 at 17:27
Michael Hardy
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answered Aug 27 "13 at 14:41
lab bhattacharjeelab bhattacharjee
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For any angle $ heta$, let $c_ heta = cos heta, s_ heta = sin heta$ and $t_ heta = an heta$, we have:

$$eginalign e^iA e^iB e^iCvà = ( c_A + i s_A )(c_B + i s_B)(c_C + is_C)\<6pt>& = c_A c_B c_C (1 + i t_A)(1 + i t_B )(1 + i t_C)\& = c_A c_B c_C igg< ig( 1 - (t_A t_B + t_B t_C + t_C t_A ) ig) + i ig( t_A + t_B + t_C - t_A t_B t_C ig)igg>endalign$$This implies $$fracIm(e^iA e^iB e^iC)Re(e^iA e^iB e^iC) = fract_A + t_B + t_C - t_A t_B t_C1 - t_A t_B - t_B t_C - t_C t_A ag*$$

One the other hand,

$$e^iA e^iB e^iC = e^i(A+B+C) = c_A+B+C(1 + i t_A+B+C),$$The L.H.S of $(*)$ is simply $t_A+B+C$. From this, we get the addition formula of tangentfor three angles:

$$t_A+B+C = fract_A + t_B + t_C - t_A t_B t_C1 - t_A t_B - t_B t_C - t_C t_A\iff an(A+B+C) = fracchảy A + chảy B + ung C - chảy A an B an C1 - an A an B - an Bchảy C - chảy Cchảy A$$In particular, this means$$chảy A + chảy B + ã C = an Achảy B ung C iff an(A+B+C) = 0$$