IPMAT Indore 2022 (SA) - If + =sqrt(2)sqrt(3) and + =1sqrt(3), then the value of (20 (-/2))^2 is _________. | PYQs + Solutions

IPMAT Indore 2022

Geometry

Trigonometry

Hard

If $\sin \alpha+\sin \beta=\frac{\sqrt{2}}{\sqrt{3}}$ and $\cos \alpha+\cos \beta=\frac{1}{\sqrt{3}}$ , then the value of $\left(20 \cos \left(\frac{\alpha-\beta}{2}\right)\right)^{2}$ is _________.

Entered answer:

✅ Correct Answer: 100

We know:

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\sin \alpha + \sin \beta = \dfrac{\sqrt{2}}{\sqrt{3}}

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\cos \alpha + \cos \beta = \dfrac{1}{\sqrt{3}}

Using trigonometric identities:

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\sin \alpha + \sin \beta = 2\sin\left(\dfrac{\alpha+\beta}{2}\right)\cos\left(\dfrac{\alpha-\beta}{2}\right)

\cos \alpha + \cos \beta = 2\cos\left(\dfrac{\alpha+\beta}{2}\right)\cos\left(\dfrac{\alpha-\beta}{2}\right)

This gives us:

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2\sin\left(\dfrac{\alpha+\beta}{2}\right)\cos\left(\dfrac{\alpha-\beta}{2}\right) = \dfrac{\sqrt{2}}{\sqrt{3}}

2\cos\left(\dfrac{\alpha+\beta}{2}\right)\cos\left(\dfrac{\alpha-\beta}{2}\right) = \dfrac{1}{\sqrt{3}}

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Dividing these equations:

\tan\left(\dfrac{\alpha+\beta}{2}\right) = \dfrac{\sin\left(\dfrac{\alpha+\beta}{2}\right)}{\cos\left(\dfrac{\alpha+\beta}{2}\right)} = \dfrac{\dfrac{\sqrt{2}}{\sqrt{3}}}{\dfrac{1}{\sqrt{3}}} = \sqrt{2}

From the second equation:

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\cos\left(\dfrac{\alpha+\beta}{2}\right)\cos\left(\dfrac{\alpha-\beta}{2}\right) = \dfrac{1}{2\sqrt{3}}

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Since

\tan\left(\dfrac{\alpha+\beta}{2}\right) = \sqrt{2}

, we get:

\cos\left(\dfrac{\alpha+\beta}{2}\right) = \dfrac{1}{\sqrt{1+\tan^2\left(\dfrac{\alpha+\beta}{2}\right)}} = \dfrac{1}{\sqrt{3}}

Substituting:

\dfrac{1}{\sqrt{3}} \cdot \cos\left(\dfrac{\alpha-\beta}{2}\right) = \dfrac{1}{2\sqrt{3}}

\cos\left(\dfrac{\alpha-\beta}{2}\right) = \dfrac{1}{2}

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Therefore:

\left(20 \cos\left(\dfrac{\alpha-\beta}{2}\right)\right)^2 = \left(20 \cdot \dfrac{1}{2}\right)^2 = 10^2 = 100