IPMAT Indore 2022 · Quantitative Ability · Question 10

Step 1 — square and add. This is the standard move whenever a sine sum and a cosine sum are both given.

{\left( \sin \alpha +\sin \beta \right)}^{2}+{\left( \cos \alpha +\cos \beta \right)}^{2}=\frac{2}{3}+\frac{1}{3}=1

Step 2 — expand the left side.

\underset{1}{\underset{⏟}{{\sin}^{2}\alpha +{\cos}^{2}\alpha}}+\underset{1}{\underset{⏟}{{\sin}^{2}\beta +{\cos}^{2}\beta}}+2\underset{\cos \left( \alpha -\beta \right)}{\underset{⏟}{\left( \sin \alpha \sin \beta +\cos \alpha \cos \beta \right)}}

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

Step 3 — solve for the difference.

2+2\cos \left( \alpha -\beta \right)=1 \Rightarrow \cos \left( \alpha -\beta \right)=-\frac{1}{2}

Step 4 — halve the angle. Using \cos 2\theta =2{\cos}^{2}\theta -1 with \theta =\frac{\alpha -\beta}{2}:

{\cos}^{2}\frac{\alpha -\beta}{2}=\frac{1+\cos \left( \alpha -\beta \right)}{2}=\frac{1-\frac{1}{2}}{2}=\frac{1}{4}

Step 5 — finish.

{\left( 20\cos \frac{\alpha -\beta}{2}\right)}^{2}=400\times \frac{1}{4}=100

Answer: 100

Shortcut. “Square and add” is the reflex whenever you are handed \sin \alpha +\sin \beta and \cos \alpha +\cos \beta. It always produces

2+2\cos \left( \alpha -\beta \right)=\left( \text{sum of the two squares}\right)

Here that is 1, so \cos \left( \alpha -\beta \right)=-1/2 and {\cos}^{2}\frac{\alpha -\beta}{2}=1/4.

Note the question asks for the square, so the sign of \cos \frac{\alpha -\beta}{2} never matters — a deliberate kindness that lets you skip any quadrant discussion. Answer 400\times 1/4=100.