Question

the value of cos255+sin195?

Answer 1

The value of $\cos 255^{\circ}+\sin 195^{\circ}\bold{=\frac{1-\sqrt{3}}{\sqrt{2}}}$.

$\cos 255^{\circ}+\sin 195^{\circ}=\cos \left(270^{\circ}-15^{\circ}\right)+\sin \left(180^{\circ}+15^{\circ}\right)$

$\begin{aligned} \cos (A-B) &=\cos A \cos B+\sin A \sin B \\ \sin (A+B) &=\sin A \cos B+\cos A \sin B \end{aligned}$

$\cos \left(270^{\circ}-15^{\circ}\right)=\left(\cos 270^{\circ} \times \cos 15^{\circ}\right)-\left(\sin 270^{\circ} \times \sin 15^{\circ}\right)$

$\sin \left(180^{\circ}+15^{\circ}\right)=\left(\sin 180^{\circ} \times \cos 15^{\circ}\right)+\left(\cos 180^{\circ} \times \sin 15^{\circ}\right)$

We know that the values of  

$\begin{aligned} \cos 270^{\circ} &=\cos \left(270^{\circ}+0^{\circ}\right)=\sin 0=0 \\ \sin 270^{\circ} &=\sin \left(270^{\circ}+0^{\circ}\right)=-\cos 0=-1 \\ \sin 180^{\circ} &=\sin \left(180^{\circ}+0^{\circ}\right)=-\sin 0=0 \\ \cos 180^{\circ} &=\cos \left(90^{\circ}+90^{\circ}\right)=-\cos 0=-1 \end{aligned}$

By substuting the above values, we can get

$\begin{aligned} \cos \left(270^{\circ}-15^{\circ}\right)=& 0 . \cos 15^{\circ}+(-1) \cdot \sin 15^{\circ}=-\sin 15^{\circ} \\ \sin \left(180^{\circ}+15^{\circ}\right) &=0 . \cos 15^{\circ}+(-1) \cdot \sin 15^{\circ} \\ &=0-\sin 15^{\circ} \end{aligned}$

By adding both the equations, we can get the  

Value of $\cos 255^{\circ}+\sin 195^{\circ}=-2 \sin 15^{\circ}$

$\begin{array}{l}{=-2 \times \frac{\sqrt{3}-1}{2 \sqrt{2}}} \\ {=\frac{1-\sqrt{3}}{\sqrt{2}}}\end{array}$

$\bold{\cos 255^{\circ}+\sin 195^{\circ}=\frac{1-\sqrt{3}}{\sqrt{2}}}$.