Question

Q.11. If $\cos (\theta -\alpha )=aand\sin (\theta -\beta )=b$, then show that ${\cos }^2\left(\alpha -\beta \right)+2ab\sin \left(\alpha -\beta \right)=a^2+b^2$

Answer 1

Dear student

$$\begin{gathered} \mathbf{Given}:\cos \left(\mathrm{\theta }-\mathrm{\alpha }\right)=\mathrm{a} \\ \mathrm{and}\sin \left(\mathrm{\theta }-\mathrm{\beta }\right)=\mathrm{b} \\ \mathrm{Consider}, \\ {\cos }^2\left(\mathrm{\alpha }-\mathrm{\beta }\right)+2\mathrm{absin}\left(\mathrm{\alpha }-\mathrm{\beta }\right) \\ ={\cos }^2\left[\left(\mathrm{\theta }-\mathrm{\beta }\right)-\left(\mathrm{\theta }-\mathrm{\alpha }\right)\right]+2\mathrm{absin}\left[\left(\mathrm{\theta }-\mathrm{\beta }\right)-\left(\mathrm{\theta }-\mathrm{\alpha }\right)\right] \\ ={\left[\cos \left(\mathrm{\theta }-\mathrm{\beta }\right)\cos \left(\mathrm{\theta }-\mathrm{\alpha }\right)+\sin \left(\mathrm{\theta }-\mathrm{\beta }\right)\sin \left(\mathrm{\theta }-\mathrm{\alpha }\right)\right]}^2+2\mathrm{ab}\left[\sin \left(\mathrm{\theta }-\mathrm{\beta }\right)\cos \left(\mathrm{\theta }-\mathrm{\alpha }\right)-\cos \left(\mathrm{\theta }-\mathrm{\beta }\right)\sin \left(\mathrm{\theta }-\mathrm{\alpha }\right)\right]\left[\mathbf{\because }\mathbf{cos}\left(\mathbf{A}\mathbf{-}\mathbf{B}\right)\mathbf{=}\mathbf{cosAcosB}\mathbf{+}\mathbf{sinAsinB}\mathbf{}\mathbf{and}\mathbf{}\mathbf{sin}\left(\mathbf{A}\mathbf{-}\mathbf{B}\right)\mathbf{=}\mathbf{sinA}\mathbf{}\mathbf{cosB}\mathbf{-}\mathbf{cosAsinB}\right] \\ ={\left[\mathrm{acos}\left(\mathrm{\theta }-\mathrm{\beta }\right)+\mathrm{bsin}\left(\mathrm{\theta }-\mathrm{\alpha }\right)\right]}^2+2\mathrm{ab}\left[\mathrm{ab}-\cos \left(\mathrm{\theta }-\mathrm{\beta }\right)\sin \left(\mathrm{\theta }-\mathrm{\alpha }\right)\right] \\ ={\mathrm{a}}^2{\cos }^2\left(\mathrm{\theta }-\mathrm{\beta }\right)+{\mathrm{b}}^2{\sin }^2\left(\mathrm{\theta }-\mathrm{\alpha }\right)+\cancel{2\mathrm{abcos}\left(\mathrm{\theta }-\mathrm{\beta }\right)\sin \left(\mathrm{\theta }-\mathrm{\alpha }\right)}+2{\mathrm{a}}^2{\mathrm{b}}^2-\cancel{2\mathrm{abcos}\left(\mathrm{\theta }-\mathrm{\beta }\right)\sin \left(\mathrm{\theta }-\mathrm{\alpha }\right)}\left[\mathbf{\because }{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{ab}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}{\mathbf{)}}^{\mathbf{2}}\right] \\ ={\mathrm{a}}^2\left(1-{\sin }^2\left(\mathrm{\theta }-\mathrm{\beta }\right)\right)+{\mathrm{b}}^2\left(1-{\cos }^2\left(\mathrm{\theta }-\mathrm{\alpha }\right)\right)+2{\mathrm{a}}^2{\mathrm{b}}^2\left[\mathbf{\because }{\mathbf{cos}}^{\mathbf{2}}\mathbf{x}\mathbf{+}{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\mathbf{=}\mathbf{1}\right] \\ ={\mathrm{a}}^2\left(1-{\mathrm{b}}^2\right)+{\mathrm{b}}^2(1-{\mathrm{a}}^2)+2{\mathrm{a}}^2{\mathrm{b}}^2 \\ ={\mathrm{a}}^2-\cancel{{\mathrm{a}}^2{\mathrm{b}}^2}+{\mathrm{b}}^2-\cancel{{\mathrm{a}}^2{\mathrm{b}}^2}+\cancel{2{\mathrm{a}}^2{\mathrm{b}}^2} \\ ={\mathrm{a}}^2+{\mathrm{b}}^2 \\ \mathrm{Hence},\boxed{{\mathbf{cos}}^{\mathbf{2}}\left(\mathbf{\alpha }\mathbf{-}\mathbf{\beta }\right)\mathbf{+}\mathbf{2}\mathbf{absin}\left(\mathbf{\alpha }\mathbf{-}\mathbf{\beta }\right)\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}} \end{gathered}$$
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