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[tex]\large\begin{array}{l} \mathtt{If~tan\,\dfrac{y}{2}=sin\,\dfrac{y}{2},~then~what~is~ cos\,y?} \end{array}[/tex]
[tex]\large\begin{array}{l} \mathtt{tan\,\dfrac{y}{2}=sin\,\dfrac{y}{2}}\\\\ \mathtt{\dfrac{sin\,\frac{y}{2}}{cos\,\frac{y}{2}}=sin\,\dfrac{y}{2}}\\\\ \mathtt{\dfrac{sin\,\frac{y}{2}}{cos\,\frac{y}{2}}-sin\,\dfrac{y}{2}=0} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{Take out the common factor }\mathtt{sin\,\dfrac{y}{2}:}\\\\ \mathtt{sin\,\dfrac{y}{2}\cdot \left(\dfrac{1}{cos\,\frac{y}{2}}-1\right)=0}\\\\ \mathtt{sin\,\dfrac{y}{2}\cdot \left(\dfrac{1}{cos\,\frac{y}{2}}-\dfrac{cos\,\frac{y}{2}}{cos\,\frac{y}{2}}\right)=0}\\\\ \mathtt{sin\,\dfrac{y}{2}\cdot \left(\dfrac{1-cos\,\frac{y}{2}}{cos\,\frac{y}{2}}\right)=0} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{If a product is zero, then one of the factors must be zero.}\\\texttt{So you must have}\\\\ \mathtt{sin\,\dfrac{y}{2}=0~~~~or~~~~\dfrac{1-cos\,\frac{y}{2}}{cos\,\frac{y}{2}}=0\qquad\qquad where~~cos\,\dfrac{y}{2}\ne 0.} \end{array}[/tex]
[tex]\large\texttt{Solving them separately:}[/tex]
• [tex]\large\begin{array}{l} \texttt{First possibility: }\mathtt{sin\,\dfrac{y}{2}=0.} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{Square both sides:}\\\\ \mathtt{sin^2\,\dfrac{y}{2}=0}\\\\\\ \texttt{Recall one of the half-angle relations:}\\\\ \mathtt{sin^2\,\dfrac{y}{2}=\dfrac{1}{2}\,(1-cos\,y)} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{Substitute back for }\mathtt{sin^2\,\dfrac{y}{2}}\texttt{ then you must have}\\\\ \mathtt{\dfrac{1}{2}\,(1-cos\,y)=0}\\\\ \mathtt{1-cos\,y=0} \end{array}[/tex]
[tex]\large\begin{array}{l} \mathtt{cos\,y=1} \end{array}[/tex] ✔
• [tex]\large\begin{array}{l} \texttt{Second possibility: }\mathtt{\dfrac{1-cos\,\frac{y}{2}}{cos\,\frac{y}{2}}=0.} \end{array}[/tex]
[tex]\large\begin{array}{l} \mathtt{1-cos\,\dfrac{y}{2}=0}\\\\ \mathtt{cos\,\dfrac{y}{2}=1} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{Square both sides:}\\\\
\mathtt{cos^2\,\dfrac{y}{2}=1}\\\\\\ \texttt{Recall other of the
half-angle relations:}\\\\
\mathtt{cos^2\,\dfrac{y}{2}=\dfrac{1}{2}\,(1+cos\,y)} \end{array}[/tex]
[tex]\large\begin{array}{l} \texttt{Substitute back for }\mathtt{cos^2\,\dfrac{y}{2}}\texttt{ then you must have}\\\\ \mathtt{\dfrac{1}{2}\,(1+cos\,y)=1}\\\\ \mathtt{1+cos\,y=2}\\\\ \mathtt{cos\,y=2-1}\end{array}[/tex]
[tex]\large\begin{array}{l} \mathtt{cos\,y=1} \end{array}[/tex] ✔
[tex]\large\texttt{So one way or another, you find}[/tex]
[tex]\large\begin{array}{l} \mathtt{cos\,y=1}\quad\longleftarrow\quad\texttt{this is the answer.} \end{array}[/tex]
[tex]\large\texttt{I hope this helps. =)}[/tex]
Tags: trigonometric trig relation half angle identity tangent sine cosine tan sin cos reduction solve equation trigonometry
