JAMIA 2023 — Mathematics PYQ

Solution

1. Use the complementary angle identity:

Recall that $\sin (\frac{\pi }{2}-\theta )=\cos \theta$. We can apply this to the last two terms:

• For $\frac{7\pi }{16}$: $\frac{\pi }{2}-\frac{\pi }{16}=\frac{8\pi -\pi }{16}=\frac{7\pi }{16}⟹\sin \frac{7\pi }{16}=\cos \frac{\pi }{16}$

• For $\frac{5\pi }{16}$: $\frac{\pi }{2}-\frac{3\pi }{16}=\frac{8\pi -3\pi }{16}=\frac{5\pi }{16}⟹\sin \frac{5\pi }{16}=\cos \frac{3\pi }{16}$

2. Substitute these into the original expression:

3. Apply the double-angle identity:

Using the identity $\sin \theta \cos \theta =\frac{1}{2}\sin 2\theta$:

• First pair: $\sin \frac{\pi }{16}\cos \frac{\pi }{16}=\frac{1}{2}\sin \left(2\times \frac{\pi }{16}\right)=\frac{1}{2}\sin \frac{\pi }{8}$

• Second pair: $\sin \frac{3\pi }{16}\cos \frac{3\pi }{16}=\frac{1}{2}\sin \left(2\times \frac{3\pi }{16}\right)=\frac{1}{2}\sin \frac{3\pi }{8}$

Now the expression is:

4. Repeat the process for the new terms:

Notice that $\frac{3\pi }{8}=\frac{\pi }{2}-\frac{\pi }{8}$, so $\sin \frac{3\pi }{8}=\cos \frac{\pi }{8}$.

Substituting this back:

Using the identity $\sin \theta \cos \theta =\frac{1}{2}\sin 2\theta$ again:

5. Final Value:

We know that $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$.

Final Answer:

The value is:

(In rationalized form, this can also be written as $\frac{\sqrt{2}}{16}$)