prove that the value of 5cosθ + 3cos(θ+π/3) + 3 lies between -4 and 10
Aditya Trivedi , 10 Years ago
Grade 10
Trigonometry> prove that the value of 5cosθ + 3cos(θ+π/...
1 AnswersY RAJYALAKSHMI
Last Activity: 10 Years ago
5cosθ + 3cos(θ+π/3) + 3 = 5cosθ + 3(cosθcosπ/3 – sinθsinπ/3) + 3
= 5cosθ + 3*cosθ*1/2 – 3*sinθ*sqrt(3/2) + 3
= 13/2*cosθ – 3sqrt(3/2)*sinθ + 3
The range of acosθ + bsinθ + c is c ± sqrt (a2 + b2)
Hence the range of the given function is 3± sqrt [(13/2)2 + (3sqrt3/2)2]
= – 4 & 10
Hence the given function lies between – 4 and 10
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