Question

# Prove that (2√3+3)sinθ+2√3cosθ lies between -(2√3+√15) and (2√3+√15).

Solution

## (2√3+3)sinθ+2√3cosθ assume a=2√3+3,b=2√3 √a2+b2=√12+9+12√3+12=√33+12√3 Dividing and multiplying the above equation with above value we get, √33+12√3(2√3+3√33+12√3sinθ+2√3√33+12√3cosθ) Assume tanϕ=ab, We have sinϕ=a√a2+b2.cosϕ=b√a2+b2 So above expression changes to √33+12√3(sinϕsinθ+cosϕcostheta) which is equal to √33+12√3cos(θ−ϕ) We know that maximum and minimum value of any cosine term is +1 and -1 √33+12√3=√15+12+6+12√3 We know that 12√3+6<12√5 because value of √5−√3 is more than 0.5 so if we replace 12√3+6 with 12√5 the above inequality still holds So range of above expression can be √15+12+12√5=2√3+√15 −(2√3+√15)<√33+12√3cos(θ−ϕ)<2√3+√15

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