Question

Please someone help me...​

Answer 1

Answer: see proof below

Explanation:

Use the following Sum Identities:

cos (A + B) = cosA · cosB - sinA · sinB

sin (A + B) = sinA · cos B - cosA · sinB

Use the Unit Circle to evaluate the following:

cos 30 = √3/2 sin 30 = 1/2

cos 45 = √2/2 sin 45 = √2/2

cos 120 = -1/2 sin 120 = √3/2

cos 240 = -1/2 sin 240 = -√3/2

cos 315 = √2/2 sin 315 = -√2/2

cos 330 = √3/2 sin 330 = -1/2

Proof LHS → RHS

`\text{LHS:}\qquad \qquad (\cos 285+\cos 345)/(\sin 435-\sin 375)`

`\text{Expand:}\qquad \quad (\cos (240+45)+\cos (315+30))/(\sin (315+120)-\sin (330+45))`

`\text{Sum Identity:}\qquad (\cos 240\cdot \cos 45-\sin 240\cdot 45+\cos 315\cdot \cos 30-\sin 315\cdot 30)/(\sin 315\cdot \cos 120+\cos315\cdot \sin 120-(\sin330\cdot \cos45+\cos 330\cdot \sin 45) )`

`\text{Unit Circle:}\quad (((-1)/(2)\cdot (\sqrt2)/(2))-((-\sqrt3)/(2)\cdot (\sqrt2)/(2))+((\sqrt2)/(2)\cdot (\sqrt3)/(3))-((-\sqrt2)/(2)\cdot (1)/(2)))/(((-\sqrt2)/(2)\cdot (-1)/(2))+((\sqrt2)/(2)\cdot (\sqrt3)/(2))-((-1)/(2)\cdot (\sqrt2)/(2))-((\sqrt3)/(2)\cdot (\sqrt2)/(2)))`

`\text{Simplify:}\qquad (-\sqrt2+\sqrt6+\sqrt6+\sqrt2)/(\sqrt2+\sqrt6+\sqrt2-\sqrt6)\qquad =(2\sqrt6)/(2\sqrt2)\qquad =\sqrt3`

LHS = RHS:
`\sqrt3 = \sqrt3\qquad \checkmark`