Question

cos48 degree ×cos12 degree

Answer 1

cos48.cos12

= 1/2 ( 2cos48.cos12)

= 1/2 ( cos(48+12) + cos ( 48-12)

= 1/2 ( cos60.cos36)

= 1/2 ( 1/2. √5+1/4)

=(√5+1 )/16

Answer 2

Answer:

$\cos (48^\circ) \cos (12^\circ)=\frac{3+\sqrt5}{8}$

Step-by-step explanation:

To find : Solve the expression $\cos (48^\circ)\times \cos (12^\circ)$

Solution :

Expression $\cos (48^\circ)\times \cos (12^\circ)$

Applying trigonometric identities,

$\cos A\cos B=\frac{1}{2}[\cos(A+B)+\cos(A-B)]$

Substitute, A=48 and B = 12

$\cos (48^\circ) \cos (12^\circ)=\frac{1}{2}[\cos(48+12)+\cos(48-12)]$

$\cos (48^\circ) \cos (12^\circ)=\frac{1}{2}[\cos(60)+\cos(36)]$

Substitute, $\cos(60)=\frac{1}{2}\ , \cos (36)=\frac{\sqrt5+1}{4}$

$\cos (48^\circ) \cos (12^\circ)=\frac{1}{2}[\frac{1}{2}+\frac{\sqrt5+1}{4}]$

$\cos (48^\circ) \cos (12^\circ)=\frac{1}{4}+\frac{\sqrt5+1}{8}$

$\cos (48^\circ) \cos (12^\circ)=\frac{2+\sqrt5+1}{8}$

$\cos (48^\circ) \cos (12^\circ)=\frac{3+\sqrt5}{8}$

Therefore, $\cos (48^\circ) \cos (12^\circ)=\frac{3+\sqrt5}{8}$