# 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 thateginequation* 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.

Bạn đang xem: Algebra precalculus  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\$\$ 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 ShobhitShobhit
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Here is a geometric proof, for the case that all three angles are acute: \$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 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