Question

Prove that sin 750° cos 480° + cos 120° cos 60° = $\frac{-1}{2}$

Answer 1

Solution :

i ) sin750°

= sin( 2×360° + 30° )

= sin 30°

= 1/2

ii ) cos 480°

= cos ( 360° + 120° )

= cos 120°

= cos ( 180 - 60° )

= - cos 60°

= -1/2

iii ) cos 120°

= cos ( 180° - 60° )

= - cos 60°

= -1/2

iv ) cos 60° = 1/2

Now ,

LHS =sin750°cos480°+cos 120°cos 60°

= ( 1/2 )( -1/2 ) + ( -1/2 )( 1/2 )

= -1/4 - 1/4

= -2/4

= -1/2

= RHS

•••••

Answer 2

Step-by-step explanation:

$\ = \\sin(750) \times \cos(480) + \cos(120) \times \cos(60) \\ \\ = \sin(360 + 390) \times \cos(360 + 120) + \cos(360 + 120) \times \cos(60) \\ = \sin(390) \times \cos(120) - \cos(60) \times \cos(60) \\ \sin(360 + 30) \times \cos(180 - 60) - \cos(60) \times \cos(60) \\ \sin(30) \times - \cos(60) - \cos(60) \times \cos(60) \\ = 1 \div 2 \times - 1 \div 2 - 1 \div 2 \times 1 \div 2 \\ = - 1 \div 4 - 1 \div 4 \\ = - 1 - 1 \div 4 \\ = - 2 \div 4 \\ = - 1 \div 4$