The value of cos248° – sin212° is ___________.
$\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}$
$=\cos ^{2} 48^{\circ}-\cos ^{2}\left(90^{\circ}-12^{\circ}\right)$
$=\cos ^{2} 48^{\circ}-\cos ^{2} 78^{\circ}$
$=(\cos 48-\cos 78)(\cos 48+\cos 78)$
$=\left(2 \sin 63^{\circ} \sin 15^{\circ}\right)\left(2 \cos 63^{\circ} \cos 15^{\circ}\right)$
Using identities :
$\cos A-\cos B=2 \sin \frac{(A+B)}{2} \sin \frac{(B-A)}{2}$ and $\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$=\left(2 \sin 63^{\circ} \cos 63^{\circ}\right)\left(2 \sin 15^{\circ} \cos 15^{\circ}\right)$
$=\sin 126^{\circ} \sin 30^{\circ}$
$=\frac{1}{2} \sin 126^{\circ}=\frac{1}{2} \sin \left(90^{\circ}+36^{\circ}\right)=\frac{1}{2} \sin 36^{\circ}$ is the answer $\left(\cdots \sin 30^{\circ}=\frac{1}{2}\right.$ and $\left.\sin \left(90^{\circ}+\theta\right)=\sin \theta\right)$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.