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Question

Prove that:
(i) cos 55° + cos 65° + cos 175° = 0

(ii) sin 50° − sin 70° + sin 10° = 0

(iii) cos 80° + cos 40° − cos 20° = 0

(iv) cos 20° + cos 100° + cos 140° = 0

(v) sin5π18-cos4π9=3 sinπ9

(vi) cosπ12-sinπ12=12

(vii) sin 80° − cos 70° = cos 50°

(viii) sin 51° + cos 81° = cos 21°

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Solution


(i)

Consider LHS:cos 55° + cos 65° + cos 175°= 2cos 55° + 65°2 cos 55° - 65°2 + cos 175° cosA+cosB=2cosA+B2cosA-B2= 2cos 60° cos-5° + cos 175°= 2×12cos 5° + cos 175°=cos 5° + cos 175°= 2cos 5° + 175°2 cos 5° - 175°2= 2cos 90° cos 85°= 0Hence, LHS=RHS.


(ii)

Consider LHS:sin 50° - sin 70° + sin 10°= 2sin 50° -70°2 cos 50° + 70°2 + sin 10° sin A - sin B = 2sin A - B2 cos A + B2= 2sin -10° cos 60° + sin 10°=2×12sin -10° + sin 10°= -sin 10°+sin 10°= 0Hence, LHS=RHS.

(iii)

Consider LHS:cos 80° + cos 40° - cos 20°= 2cos 80° + 40°2 cos 80° - 40°2 - cos 20° cosA+cosB=2cosA+B2cosA-B2= 2cos 60° cos 20°- cos 20°= 2×12cos 20° - cos20°= cos 20° - cos 20°= 0Hence, LHS=RHS.

(iv)

Consider LHS:cos 20° + cos 100° + cos 140°= 2cos 20° + 100°2 cos 20° - 100°2 + cos 140° cosA+cosB=2cosA+B2cosA-B2= 2cos 60° cos -40° + cos 140°= 2×12cos 40° + cos 140°= cos 40° + cos 140°= 2cos 40° + 140°2 cos 40° - 140°2= 2 cos 90° cos 50°= 0Hence, LHS=RHS.


(v)

LHS=sin5π18-cos4π9=sin5π18- cosπ2-π18=sin5π18- sinπ18=2sin5π18-π182cos5π18+π182 sinA-sinB=2sinA-B2cosA+B2=2sinπ9cosπ6=2sinπ9cosπ6=2×32sinπ9=3sinπ9=RHSHence, LHS=RHS.

(vi)

LHS= cos π12 - sin π12= cos π2 - 5π12 - sin π12= sin 5π12 - sin π12= 2sin 5π12 - π122 cos 5π12 + π122 sin A - sin B = 2sin A - B2 cos A + B2= 2sin π6 cos π4= 2×12×12=12Hence, LHS=RHS.


(vii)

Consider LHS:sin 80° - cos 70°= sin 80° - cos 90° - 20°= sin 80° - sin 20°= 2sin 80° - 20°2 cos 80° + 20°2 sin A - sin B = 2sin A - B2 cos A + B2= 2sin 30° cos 50°= 2×12cos 50°= cos 50°= RHSHence, LHS=RHS.

(viii)

Consider LHS:sin 51° + cos 81°= sin 51° + cos 90° - 9°= sin 51° + sin 9°= 2sin 51° + 9°2 cos 51° - 9°2 sin A + sin B = 2sin A + B2 cos A - B2= 2sin 30° cos 21°= 2×12cos21°= cos21°= RHSHence, LHS=RHS.

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