proof of double angle identity
ProofOfDoubleAngleIdentity
2002-07-15 22:27:09
2002-07-15 22:28:57
Proof
double angle identity
0
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\textbf{Sine:}\\
\begin{eqnarray*}
\sin(2a)&=&\sin(a+a)\\
&=&\sin(a)\cos(a)+\cos(a)\sin(a)\\
&=&2\sin(a)\cos(a).
\end{eqnarray*}
\textbf{Cosine:}
\begin{eqnarray*}
\cos(2a)&=&\cos(a+a)\\
&=&\cos(a)\cos(a)+\sin(a)\sin(a)\\
&=&\cos^2(a)-\sin^2(a).
\end{eqnarray*}
By using the identity $$\sin^2(a)+\cos^2(a)=1$$ we can change the expression above into the alternate forms
$$\cos(2a)=2\cos^2(a)-1 = 1-2\sin^2(a).$$
\textbf{Tangent:}
\begin{eqnarray*}
\tan(2a)&=&\tan(a+a)\\
&=&\frac{\tan(a)+\tan(a)}{1-\tan(a)\tan(a)}\\
&=&\frac{2\tan(a)}{1-\tan^2(a)}.
\end{eqnarray*}