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Trigonometric Functions and Equations question

2022 · Shift 2 · Q19
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  5. /2022 · Shift 2 · Q19

Trigonometric Functions and Equations question

2022 · Shift 2 · Q19

JEE AdvancedMathematicsTrigonometric Functions and EquationsNumerical+3 / −1
Let α\alphaα and β\betaβ be real numbers such that −π4<β<0<α<π4-\frac{\pi}{4}\lt \beta\lt 0\lt \alpha\lt \frac{\pi}{4}−4π​<β<0<α<4π​. If sin⁡(α+β)=13\sin (\alpha+\beta)=\frac{1}{3}sin(α+β)=31​ and cos⁡(α−β)=23\cos (\alpha-\beta)=\frac{2}{3}cos(α−β)=32​, then the greatest integer less than or equal to (sin⁡αcos⁡β+cos⁡βsin⁡α+cos⁡αsin⁡β+sin⁡βcos⁡α)2\left(\frac{\sin \alpha}{\cos \beta}+\frac{\cos \beta}{\sin \alpha}+\frac{\cos \alpha}{\sin \beta}+\frac{\sin \beta}{\cos \alpha}\right)^{2}(cosβsinα​+sinαcosβ​+sinβcosα​+cosαsinβ​)2 is
Numerical answer
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Correct answer: 1

  1. Given data

We have - rac{\pi}{4}<\beta<0<\alpha<\frac{\pi}{4}, so:

  • sin⁡α>0, cos⁡α>0\sin\alpha>0,\ \cos\alpha>0sinα>0, cosα>0
  • sin⁡β<0, cos⁡β>0\sin\beta<0,\ \cos\beta>0sinβ<0, cosβ>0

Also, sin⁡(α+β)=13,cos⁡(α−β)=23.\sin(\alpha+\beta)=\frac13,\qquad \cos(\alpha-\beta)=\frac23.sin(α+β)=31​,cos(α−β)=32​.

We need to find (sin⁡αcos⁡β+cos⁡βsin⁡α+cos⁡αsin⁡β+sin⁡βcos⁡α)2\left(\frac{\sin \alpha}{\cos \beta}+\frac{\cos \beta}{\sin \alpha}+\frac{\cos \alpha}{\sin \beta}+\frac{\sin \beta}{\cos \alpha}\right)^2(cosβsinα​+sinαcosβ​+sinβcosα​+cosαsinβ​)2 and then its floor.

  1. Introduce simpler variables

Let x=sin⁡αcos⁡β,y=sin⁡βcos⁡α.x=\frac{\sin\alpha}{\cos\beta},\qquad y=\frac{\sin\beta}{\cos\alpha}.x=cosβsinα​,y=cosαsinβ​. Then x+y=sin⁡αcos⁡α?x+y=\frac{\sin\alpha\cos\alpha?}{ }x+y=sinαcosα?​ But more directly,

=\frac{\sin\alpha\cos\alpha+\sin\beta\cos\beta}{\cos\alpha\cos\beta}$$ which is not immediately helpful. A better observation is: $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.$$ Dividing by $\cos\alpha\cos\beta$ gives $$\tan\alpha+\tan\beta=\frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta}.$$ Similarly, $$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.

But the target expression simplifies neatly if we write it as E=sin⁡αcos⁡β+cos⁡βsin⁡α+cos⁡αsin⁡β+sin⁡βcos⁡α.E=\frac{\sin \alpha}{\cos \beta}+\frac{\cos \beta}{\sin \alpha}+\frac{\cos \alpha}{\sin \beta}+\frac{\sin \beta}{\cos \alpha}.E=cosβsinα​+sinαcosβ​+sinβcosα​+cosαsinβ​. Group terms: E=(sin⁡αcos⁡β+sin⁡βcos⁡α)+(cos⁡βsin⁡α+cos⁡αsin⁡β).E=\left(\frac{\sin\alpha}{\cos\beta}+\frac{\sin\beta}{\cos\alpha}\right)+\left(\frac{\cos\beta}{\sin\alpha}+\frac{\cos\alpha}{\sin\beta}\right).E=(cosβsinα​+cosαsinβ​)+(sinαcosβ​+sinβcosα​).

Now,

=\frac{\sin\alpha\cos\alpha+\sin\beta\cos\beta}{\cos\alpha\cos\beta},$$ and $$\frac{\cos\beta}{\sin\alpha}+\frac{\cos\alpha}{\sin\beta} =\frac{\sin\beta\cos\beta+\sin\alpha\cos\alpha}{\sin\alpha\sin\beta}.$$ So if we define $$S=\sin\alpha\cos\alpha+\sin\beta\cos\beta,$$ then $$E=S\left(\frac{1}{\cos\alpha\cos\beta}+\frac{1}{\sin\alpha\sin\beta}\right).

Thus

Using identities, sin⁡αsin⁡β+cos⁡αcos⁡β=cos⁡(α−β)=23.\sin\alpha\sin\beta+\cos\alpha\cos\beta=\cos(\alpha-\beta)=\frac23.sinαsinβ+cosαcosβ=cos(α−β)=32​. Also,

Hence

So

Now, 4sin⁡αsin⁡βcos⁡αcos⁡β=(2sin⁡αcos⁡α)(2sin⁡βcos⁡β)=sin⁡2αsin⁡2β.4\sin\alpha\sin\beta\cos\alpha\cos\beta=(2\sin\alpha\cos\alpha)(2\sin\beta\cos\beta)=\sin2\alpha\sin2\beta.4sinαsinβcosαcosβ=(2sinαcosα)(2sinβcosβ)=sin2αsin2β. Thus

  1. Find sin⁡2αsin⁡2β\sin2\alpha\sin2\betasin2αsin2β

We know sin⁡(α+β)=13.\sin(\alpha+\beta)=\frac13.sin(α+β)=31​. Since −π4<β<0<α<π4-\frac\pi4<\beta<0<\alpha<\frac\pi4−4π​<β<0<α<4π​, we have −π4<α+β<π4,-\frac\pi4<\alpha+\beta<\frac\pi4,−4π​<α+β<4π​, so α+β∈(−π4,π4)\alpha+\beta\in\left(-\frac\pi4,\frac\pi4\right)α+β∈(−4π​,4π​) and given sin⁡(α+β)=13>0\sin(\alpha+\beta)=\frac13>0sin(α+β)=31​>0, therefore cos⁡(α+β)=1−19=223.\cos(\alpha+\beta)=\sqrt{1-\frac19}=\frac{2\sqrt2}{3}.cos(α+β)=1−91​​=322​​.

Also, cos⁡(α−β)=23.\cos(\alpha-\beta)=\frac23.cos(α−β)=32​. Since 0<α−β<π20<\alpha-\beta<\frac\pi20<α−β<2π​, we get sin⁡(α−β)=1−49=53.\sin(\alpha-\beta)=\sqrt{1-\frac49}=\frac{\sqrt5}{3}.sin(α−β)=1−94​​=35​​.

Now use

=\frac{\cos2(\alpha-\beta)-\cos2(\alpha+\beta)}{2}.$$ Compute: $$\cos2(\alpha-\beta)=2\cos^2(\alpha-\beta)-1=2\left(\frac23\right)^2-1=\frac89-1=-\frac19.$$ And $$\cos2(\alpha+\beta)=1-2\sin^2(\alpha+\beta)=1-2\left(\frac13\right)^2=1-\frac29=\frac79.$$ Hence $$\sin2\alpha\sin2\beta=\frac{-1/9-7/9}{2}=\frac{-8/9}{2}=-\frac49.$$ --- 4. **Compute $E$** Substitute into $$E=\frac{16}{27\sin2\alpha\sin2\beta}:$$ $$E=\frac{16}{27\cdot(-4/9)}=\frac{16}{-12}=-\frac43.$$ Therefore, $$E^2=\left(-\frac43\right)^2=\frac{16}{9}.$$ So the greatest integer less than or equal to this is $$\left\lfloor \frac{16}{9}\right\rfloor=1.$$ --- 5. **Final answer** $$\boxed{1}$$ This matches the stored correct answer.
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