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Trigonometric Ratio and Identites question

2019 · 12 Jan · Shift 1 · Q36
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Trigonometric Ratio and Identites question

2019 · 12 Jan · Shift 1 · Q36

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
The maximum value of 3cos θ\thetaθ+ 5sin (θ−π6)\left( {\theta - {\pi \over 6}} \right)(θ−6π​) for any real value of θ\thetaθ is :
  1. A
    34\sqrt {34}34​
  2. B
    31\sqrt {31}31​
  3. C
    19\sqrt {19}19​
  4. D
    792{{\sqrt {79} } \over 2}279​​
View written solutionFree

Correct answer: C

  1. Given expression

We need the maximum value of 3cos⁡θ+5sin⁡(θ−π6).3\cos\theta + 5\sin\left(\theta - \frac{\pi}{6}\right).3cosθ+5sin(θ−6π​).

  1. Expand the sine term

Using sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β,\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta,sin(α−β)=sinαcosβ−cosαsinβ, we get sin⁡(θ−π6)=sin⁡θ⋅32−cos⁡θ⋅12.\sin\left(\theta-\frac{\pi}{6}\right)=\sin\theta\cdot\frac{\sqrt3}{2}-\cos\theta\cdot\frac12.sin(θ−6π​)=sinθ⋅23​​−cosθ⋅21​.

So, 5sin⁡(θ−π6)=532sin⁡θ−52cos⁡θ.5\sin\left(\theta-\frac{\pi}{6}\right)=\frac{5\sqrt3}{2}\sin\theta-\frac{5}{2}\cos\theta.5sin(θ−6π​)=253​​sinθ−25​cosθ.

Hence the expression becomes

= 3\cos\theta + \frac{5\sqrt3}{2}\sin\theta - \frac{5}{2}\cos\theta.$$ Combine the cosine terms: $$= \left(3-\frac52\right)\cos\theta + \frac{5\sqrt3}{2}\sin\theta = \frac12\cos\theta + \frac{5\sqrt3}{2}\sin\theta.$$ 3. **Use the standard maximum formula** For an expression of the form $$a\cos\theta+b\sin\theta,$$ its maximum value is $$\sqrt{a^2+b^2}.$$ Here, $$a=\frac12, \qquad b=\frac{5\sqrt3}{2}.$$ Therefore, $$\text{Maximum} = \sqrt{\left(\frac12\right)^2+\left(\frac{5\sqrt3}{2}\right)^2}.$$ Compute: $$\left(\frac12\right)^2=\frac14,$$ $$\left(\frac{5\sqrt3}{2}\right)^2=\frac{75}{4}.$$ So, $$\text{Maximum} = \sqrt{\frac14+\frac{75}{4}} = \sqrt{\frac{76}{4}} = \sqrt{19}.$$ 4. **Match with options** $$\sqrt{19}$$ corresponds to **Option C**. 5. **Compare with stored correct answer** Stored correct answer: **C** Our derived answer: **C** So, the stored answer is correct.
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