Question

Prove that : $\sin 105°+\cos 105°=\cos 45°.$

Answer 1

Use the trigonometry identity:

$\sin 105°+\cos 105°=\cos 45°.$

Taking LHS $\sin 105°+\cos 105°=\cos 45°.$

Firstly $\sin 105°=\sin (60+45)°$

Also $\cos 105°=\cos (60+45)°$

As we know that $\sin (A+B)=\sin A\cos B+\cos A\sin B$

Also, $\cos (A+B)=\cos A\cos B–\sin A\sin B$

$$\begin{gathered} \sin 105°+\cos 105° \\ =\sin (60+45)°+\cos (60+45)° \\ =\sin 60°\cos 45°+\cos 60°\sin 45°+\cos 60°\cos 45°–\sin 60°\sin 45° \\ =\cos 45°(\sin 60°+\cos 60°)+\sin 45°(\cos 60°–\sin 60°) \\ =\cos 45°(\sin 60°+\cos 60°+\cos 60°–\sin 60°)\mathbf{}\mathbf{(}\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{45}\mathbf{°}\mathbf{=}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{45}\mathbf{°}\mathbf{)} \\ =\cos 45°(2\cos 60°) \\ =\cos 45°\left(2\times \frac{1}{2}\right) \\ =\cos 45°. \end{gathered}$$

Hence proved.