Given f(θ)=3cosθ+5sin(θ−6π)
⇒f(θ)=3cosθ+5(sinθ⋅23−cosθ⋅21)
⇒f(θ)=253sinθ+21cosθ
Now using the concept acosx+bsinx+c∈[c−a2+b2,c+a2+b2], we can write
Maximum value of f(θ) is (253)2+(21)2=475+41=19
JEE Main 2019 — Mathematics Trigonometry
The maximum value of 3cosθ+5sin(θ−6π) for any real value of θ is :
Held on 12 Jan 2019 · Verified 6 Jul 2026.
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