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Transformation Formulae — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae1 Mark
Question
If $\text{A}+\text{B}=\frac{\pi}{3}\text{ and }\cos\text{A}+\cos\text{B}=1,$ than find the value of $\cos\frac{\text{A}-\text{B}}{2}.$
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Answer

We have, $\text{A+B}=\frac{\pi}{3}$ $\text{and },\cos\text{A}+\cos\text{B}=1$ Now, $\cos\text{A}+\cos\text{B}=1$ $\Rightarrow\ 2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=1$ $\Rightarrow\ 2\cos\Big(\frac{1}{2}\times\frac{\pi}{3}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=1$ $\Big[\because\ \text{A+B}=\frac{\pi}{3}\Big]$ $\Rightarrow\ 2\cos\frac{\pi}{6}\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=1$ $\Rightarrow\ 2\times\frac{\sqrt3}{2}\times\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=1$ $\Rightarrow\ \sqrt3\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=1$ $\Rightarrow\ \cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=\frac{1}{\sqrt3}$ $\text{Hence},\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)=\frac{1}{\sqrt3}.$

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