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Trigonometric Functions and Equations question

2010 · Shift 1 · Q31
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  5. /2010 · Shift 1 · Q31

Trigonometric Functions and Equations question

2010 · Shift 1 · Q31

JEE AdvancedMathematicsTrigonometric Functions and EquationsNumerical+4 / −1
The maximum value of the expression 1sin⁡2θ+3sin⁡θcos⁡θ+5cos⁡2θ{1 \over {{{\sin }^2}\theta + 3\sin \theta \cos \theta + 5{{\cos }^2}\theta }}sin2θ+3sinθcosθ+5cos2θ1​ is
Numerical answer
View written solutionFree

Correct answer: 2

  1. We need to maximize
1sin⁡2θ+3sin⁡θcos⁡θ+5cos⁡2θ.\frac{1}{\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta}.sin2θ+3sinθcosθ+5cos2θ1​.

Since the numerator is constant and positive, this expression is maximum when the denominator is minimum.

So let

D=sin⁡2θ+3sin⁡θcos⁡θ+5cos⁡2θ.D=\sin^2\theta+3\sin\theta\cos\theta+5\cos^2\theta.D=sin2θ+3sinθcosθ+5cos2θ.

We need to find the minimum value of DDD.

  1. Write sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\thetasin2θ=1−cos2θ:
D=(1−cos⁡2θ)+3sin⁡θcos⁡θ+5cos⁡2θ=1+4cos⁡2θ+3sin⁡θcos⁡θ.D=(1-\cos^2\theta)+3\sin\theta\cos\theta+5\cos^2\theta =1+4\cos^2\theta+3\sin\theta\cos\theta.D=(1−cos2θ)+3sinθcosθ+5cos2θ=1+4cos2θ+3sinθcosθ.

A more standard way is to use double-angle identities:

sin⁡2θ=1−cos⁡2θ2,cos⁡2θ=1+cos⁡2θ2,sin⁡θcos⁡θ=sin⁡2θ2.\sin^2\theta=\frac{1-\cos2\theta}{2}, \qquad \cos^2\theta=\frac{1+\cos2\theta}{2}, \qquad \sin\theta\cos\theta=\frac{\sin2\theta}{2}.sin2θ=21−cos2θ​,cos2θ=21+cos2θ​,sinθcosθ=2sin2θ​.

Substitute:

D=1−cos⁡2θ2+3⋅sin⁡2θ2+5⋅1+cos⁡2θ2.D=\frac{1-\cos2\theta}{2}+3\cdot\frac{\sin2\theta}{2}+5\cdot\frac{1+\cos2\theta}{2}.D=21−cos2θ​+3⋅2sin2θ​+5⋅21+cos2θ​.

Simplifying,

D=1−cos⁡2θ+3sin⁡2θ+5+5cos⁡2θ2=6+4cos⁡2θ+3sin⁡2θ2.D=\frac{1-\cos2\theta+3\sin2\theta+5+5\cos2\theta}{2} =\frac{6+4\cos2\theta+3\sin2\theta}{2}.D=21−cos2θ+3sin2θ+5+5cos2θ​=26+4cos2θ+3sin2θ​.

Hence,

D=3+2cos⁡2θ+32sin⁡2θ.D=3+2\cos2\theta+\frac{3}{2}\sin2\theta.D=3+2cos2θ+23​sin2θ.
  1. Now use the standard result:
acos⁡x+bsin⁡xa\cos x+b\sin xacosx+bsinx

has minimum value

−a2+b2.-\sqrt{a^2+b^2}.−a2+b2​.

Here,

a=2,b=32.a=2,\qquad b=\frac{3}{2}.a=2,b=23​.

So

a2+b2=4+94=254=52.\sqrt{a^2+b^2}=\sqrt{4+\frac{9}{4}}=\sqrt{\frac{25}{4}}=\frac{5}{2}.a2+b2​=4+49​​=425​​=25​.

Therefore the minimum value of DDD is

Dmin⁡=3−52=12.D_{\min}=3-\frac{5}{2}=\frac{1}{2}.Dmin​=3−25​=21​.
  1. Therefore the maximum value of the original expression is
1Dmin⁡=11/2=2.\frac{1}{D_{\min}}=\frac{1}{1/2}=2.Dmin​1​=1/21​=2.
  1. Final answer:
2\boxed{2}2​
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