Question

sin5°-sin67°+sin77°-sin139°+sin149°=0​

Answer 1

LHS = sin5° - sin67° + sin77° - sin139° + sin149°

we know, sinA - sinB = 2cos(A + B)/2.sin(A - B)/2

so, -sin67° +sin77° = 2cos(67°+77°)/2.sin(77° - 67°)/2

= 2cos72° . sin5°

similarly, -sin139° . sin149° = 2cos(139°+149°)/2.sin(149° - 139°)/2

= 2cos(144°).sin5°

now, LHS = sin5° + 2cos72°sin5° + 2cos144°.sin5°

= sin5° [ 1 + 2cos72° + 2cos144°]

= sin5°[1 + 2(cos72° + cos144°)]

we know, cosA + cosD = 2cos(C + D)/2.cos(C - D)/2

= sin5° [ 1 + 2{2cos(72°+144°)/2.cos(144°-72°)/2]

= sin5° [1 + 4cos108°.cos36°]

= sin5° [1 + 4cos(90° +18°) .cos36°]

= sin5° [ 1 - 4sin18°.cos36° ]

[ cos36° = (√5 + 1)/4 and sin18° = (√5 - 1)/4

so, sin18°.cos36° = (√5² - 1²)/16 = 4/16 = 1/4 ]

= sin5° [1 - 4 × 1/4 ]

= sin5° [1 - 1]

= 0 = RHS

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