Question

Express $\sin \left(67°\right)+\cos \left(75°\right)$ in terms of trigonometric ratios of angles between $0°$ and $45°$.

Answer 1

Express $\mathbf{sin}\mathbf{(}\mathbf{67}\mathbf{°}\mathbf{)}\mathbf{+}\mathbf{cos}\mathbf{(}\mathbf{75}\mathbf{°}\mathbf{)}$ in terms of trigonometric ratios of angles between $\mathbf{0}\mathbf{°}$ and $\mathbf{45}\mathbf{°}$.

Since, $67°=90°-23°$

And, $75°=90°-15°$

So, we can write that

$\sin (67°)+\cos (75°)=\sin (90°-23°)+\cos (90°-15°)$

$=\cos \left(23°\right)+\sin \left(15°\right)$ [ $\cos \left(90°-\theta \right)=\sin \theta$ and $\sin \left(90°-\theta \right)=\cos \theta$ ]

Hence, $\mathbf{sin}\mathbf{(}\mathbf{67}\mathbf{°}\mathbf{)}\mathbf{+}\mathbf{cos}\mathbf{(}\mathbf{75}\mathbf{°}\mathbf{)}$ can be expressed as $\mathbf{cos}\mathbf{(}\mathbf{23}\mathbf{°}\mathbf{)}\mathbf{+}\mathbf{sin}\mathbf{(}\mathbf{15}\mathbf{°}\mathbf{)}$.