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Difference between revisions of "Angle Addition Formulas (Trigonometry)"
Line 19: Line 19:
 
label("$\alpha$", A ,7*dir(a*30));
 
label("$\alpha$", A ,7*dir(a*30));
 
label("$\beta$", A ,9*dir(b*45+ a*30));
 
label("$\beta$", A ,9*dir(b*45+ a*30));
 +
label("$A$",A,SW);
 +
label("$B$",B,SE);
 +
label("$C$",C,E);
 +
label("$D$",D,N);
 +
label("$E$",E,S);
 +
label("$F$",F,W);
 +
draw(rightanglemark(F,E,A,1));
 +
draw(rightanglemark(C,F,D,1));
 
</asy></center>
 
</asy></center>
 +
 +
===Sine Angle Addition===
 +
We let <math>\angle BAC=\alpha</math>, <math>\angle CAD=\beta</math>, <math>E</math> the foot of the altitude from <math>D</math> to <math>AB</math> and <math>F</math> the foot of the altitude from <math>C</math> to <math>DE</math>. We let <math>AD=1</math>. Then, we have that <math>CD=\sin\beta</math> and <math>AC=\cos\beta</math>. Furthermore, we see that <math>\angle CDF=\angle CAB=\alpha</math>. Thus, we see that <math>FE=BC=AC\sin\alpha=\cos\beta\sin\alpha</math> and <math>DF=CD\cos\alpha=\cos\alpha\sin\beta</math>. Thus, we see that <math>\sin(\alpha+\beta)=DE=DF+FE=\cos\beta\sin\alpha+\sin\beta\cos\alpha</math>, giving<cmath>\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha.</cmath>
 +
 +
===Cosine Angle Addition===
 +
We let <math>\angle BAC=\alpha</math>, <math>\angle CAD=\beta</math>, <math>E</math> the foot of the altitude from <math>D</math> to <math>AB</math> and <math>F</math> the foot of the altitude from <math>C</math> to <math>DE</math>. We let <math>AD=1</math>. Then, we have that <math>CD=\sin\beta</math> and <math>AC=\cos\beta</math>. Furthermore, we see that <math>\angle CDF=\angle CAB=\alpha</math>. Thus, we see that <math>AB=AC\cos\alpha=\cos\alpha\cos\beta</math> and <math>BE=CF=CD\sin\alpha=\sin\beta\sin\alpha</math>. Thus, we see that <math>\sin(\alpha+\beta)=AE=AB-BE=\cos\beta\cos\alpha-\sin\beta\sin\alpha</math>, giving<cmath>\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\beta\sin\alpha.</cmath>
 +
 +
===Tangent Angle Addition===
 +
We have already found <math>AE</math> and <math>DE</math> above; thus we get <math>\tan(\alpha+\beta)=\frac{\sin\alpha\cos\beta+\sin\beta\cos\alpha}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}</math>. Dividing numerator and denominator by <math>\cos\alpha\cos\beta</math>, we get<cmath>\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}.</cmath>

Revision as of 18:24, 2 February 2020

[asy] real a,b,c; a=0.4924; b=0.5467; c=a+b; pair A,B,C,D,E,F; A=(0,0); D=(cos(c),sin(c)); B=(cos(a)*cos(b),0); C=(cos(a)*cos(b),sin(a)*cos(b)); draw(A--B--C--D--cycle); draw(A--C); E=foot(D,A,B); draw(D--E); F=foot(C,D,E); draw(C--F); draw(anglemark(B,A,C,4)); draw(anglemark(C,A,D,5.5)); label("$\alpha$", A ,7*dir(a*30)); label("$\beta$", A ,9*dir(b*45+ a*30)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,E); label("$D$",D,N); label("$E$",E,S); label("$F$",F,W); draw(rightanglemark(F,E,A,1)); draw(rightanglemark(C,F,D,1)); [/asy]

Sine Angle Addition

We let $\angle BAC=\alpha$, $\angle CAD=\beta$, $E$ the foot of the altitude from $D$ to $AB$ and $F$ the foot of the altitude from $C$ to $DE$. We let $AD=1$. Then, we have that $CD=\sin\beta$ and $AC=\cos\beta$. Furthermore, we see that $\angle CDF=\angle CAB=\alpha$. Thus, we see that $FE=BC=AC\sin\alpha=\cos\beta\sin\alpha$ and $DF=CD\cos\alpha=\cos\alpha\sin\beta$. Thus, we see that $\sin(\alpha+\beta)=DE=DF+FE=\cos\beta\sin\alpha+\sin\beta\cos\alpha$, giving\[\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha.\]

Cosine Angle Addition

We let $\angle BAC=\alpha$, $\angle CAD=\beta$, $E$ the foot of the altitude from $D$ to $AB$ and $F$ the foot of the altitude from $C$ to $DE$. We let $AD=1$. Then, we have that $CD=\sin\beta$ and $AC=\cos\beta$. Furthermore, we see that $\angle CDF=\angle CAB=\alpha$. Thus, we see that $AB=AC\cos\alpha=\cos\alpha\cos\beta$ and $BE=CF=CD\sin\alpha=\sin\beta\sin\alpha$. Thus, we see that $\sin(\alpha+\beta)=AE=AB-BE=\cos\beta\cos\alpha-\sin\beta\sin\alpha$, giving\[\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\beta\sin\alpha.\]

Tangent Angle Addition

We have already found $AE$ and $DE$ above; thus we get $\tan(\alpha+\beta)=\frac{\sin\alpha\cos\beta+\sin\beta\cos\alpha}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}$. Dividing numerator and denominator by $\cos\alpha\cos\beta$, we get\[\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}.\]

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