Question

The value of $1+\cos {56}^0+\cos {58}^0-\cos {66}^0$ is equal to

• A. $2$ $\cos {28}^{}\cos {29}^{}\cos {33}^{}$
• B. $4\cos {28}^0\cos {29}^0\sin {33}^0$
• C. $4\cos {28}^0\cos {29}^{}\cos {33}^{}$
• D. 2 $\cos {28}^0\cos {29}^0\sin {33}^0$

Answer 1

The correct option is B $4\cos {28}^0\cos {29}^0\sin {33}^0$

$1+\cos {56}^o+\cos {58}^o-\cos {66}^o$
$=1+\cos {56}^o+[-2\sin ({62}^o\left)\sin \right(-4\left)\right]$ .....$[\because \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)]$
$=1+\cos {56}^o+2\sin {62}^o\sin 4$ .....$[\because \sin \theta =\cos ({90}^o-\theta \left)\right]$
$=1+{\cos }^{2}28-{\sin }^{2}28+2\cos 28\sin 4$
$=2{\cos }^{2}28+2\cos 28\sin 4$ ..... $[\because {\cos }^2{28}^o+{\sin }^2{28}^o=1]$
$=2\cos {28}^o[\cos {28}^o+\sin 4^o]$
$=2\cos {28}^o[\cos {28}^o+\cos {86}^o]$ ..... $[\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)]$
$=2\cos {28}^o\left[2\cos {57}^o\cos {29}^o\right]$
$=4\cos {28}^o\cos {29}^o\cos {57}^o$
$=4\cos {28}^o\cos {29}^o\sin {33}^o$ .....$[\because \sin \theta =\cos (90-\theta \left)\right]$