Question

Evaluate$\cos (\mathrm{\pi }-\mathrm{x})$

Answer 1

We know that $\cos (\mathrm{A}-\mathrm{B})=\cos \mathrm{A}\cos \mathrm{B}+\sin \mathrm{A}\sin \mathrm{B}$

⇒ $\cos (\mathrm{\pi }-\mathrm{x})$ $=$$\cos \mathrm{\pi }\cos \mathrm{x}+\sin \mathrm{\pi }\sin \mathrm{x}$

Solving $\cos \mathrm{\pi }$,$\mathrm{sin\pi }$using supplementary angle concept of trigonometry, we get

$\cos \mathrm{\pi }$ $=$$\cos {180}^0$

$=\cos ({90}^0+{90}^0)$

$=-\sin {90}^0$

⇒$\cos \mathrm{\pi }$ $=-1$ From Trigonometric Angle Table $\sin {90}^0=1$

$\sin \mathrm{\pi }$ $=$$\sin {180}^0$

$=\sin ({90}^0+{90}^0)$

$=\cos {90}^0$

⇒$\sin \mathrm{\pi }$ $=0$ From Trigonometric Angle Table $\cos {90}^0=0$

⇒$\cos (\mathrm{\pi }-\mathrm{x})$ $=$$(-1)\times \cos \mathrm{x}+0\times \sin \mathrm{x}$ From above $\cos \mathrm{\pi }$$=-1$

$=-\cos \mathrm{x}$

Thus,$\cos (\mathrm{\pi }-\mathrm{x})$ $=-\cos \mathrm{x}$