Question

find the value of sin(45-tita )-cos (45-tita)

Answer 1

$$\begin{gathered} \sin \left(45-\mathrm{\theta }\right)-\cos \left(45-\mathrm{\theta }\right) \\ =\left(\sin 45\mathrm{cos\theta }-\cos 45\mathrm{sin\theta }\right)-\left(\cos 45\mathrm{cos\theta }+\sin 45\mathrm{sin\theta }\right)\left[\mathrm{Using},\sin \left(\mathrm{A}-\mathrm{B}\right)=\mathrm{sinAcosB}-\mathrm{cosAsinB}\mathrm{and}\cos \left(\mathrm{A}-\mathrm{B}\right)=\mathrm{cosAcosB}+\mathrm{sinAsinB}\right] \\ =\left(\frac{1}{\sqrt{2}}\mathrm{cos\theta }-\frac{1}{\sqrt{2}}\mathrm{sin\theta }\right)-\left(\frac{1}{\sqrt{2}}\mathrm{cos\theta }+\frac{1}{\sqrt{2}}\mathrm{sin\theta }\right) \\ =\frac{1}{\sqrt{2}}\mathrm{cos\theta }-\frac{1}{\sqrt{2}}\mathrm{sin\theta }-\frac{1}{\sqrt{2}}\mathrm{cos\theta }-\frac{1}{\sqrt{2}}\mathrm{sin\theta } \\ =-\frac{2}{\sqrt{2}}\mathrm{sin\theta } \\ =-\sqrt{2}\mathrm{sin\theta } \\ \Rightarrow \mathbf{sin}\left(\mathbf{45}\mathbf{-}\mathbf{\theta }\right)\mathbf{-}\mathbf{cos}\left(\mathbf{45}\mathbf{-}\mathbf{\theta }\right)\mathbf{=}\mathbf{-}\sqrt{\mathbf{2}}\mathbf{}\mathbf{sin\theta } \end{gathered}$$