Question

Show that:$\cos 38°\cos 52°–\sin 38°\sin 52°=0$

Answer 1

To prove :$\cos 38°\cos 52°–\sin 38°\sin 52°=0$

$\mathrm{L}.\mathrm{H}.\mathrm{S}=$$\cos 38°\cos 52°–\sin 38°\sin 52°$

$=$$\cos (90°-{52}^0)\cos 52°–\sin (90°-{52}^0)\sin {52}^0$ [Identity used: $\cos ({90}^0-\mathrm{\theta })=\mathrm{sin\theta }$ and $\sin ({90}^0-\mathrm{\theta })=\mathrm{cos\theta }$]

$=$$\sin {52}^0\cos 52°–\cos 52°\sin {52}^0$

$=0$

⇒ $\mathrm{L}.\mathrm{H}.\mathrm{S}=$$\mathrm{R}.\mathrm{H}.\mathrm{S}$

Thus,$\cos 38°\cos 52°–\sin 38°\sin 52°=0$. Hence proved