If L = sin²(π/16) - sin²(π/8) and M = cos²(π/16) - sin²(π/8), then (A) L = -1/(2√2) + 1/2 cos(π/8) (

Q: Wait, why do we choose to focus on $M$ first instead of $L$?

That is a great strategic question! In competitive exams like JEE, time is your most valuable resource. If you look at the options, two of them explicitly define $M$. By solving for $M$ first, you have a 50% chance of finding the correct answer immediately. It is a smart, time-saving tactic to analyze the options before diving into the calculations.

Q: Is $\cos(-\theta) = \cos(\theta)$ always true?

Yes, absolutely! This is because the cosine function is an even function. Geometrically, if you rotate an angle $\theta$ clockwise or counter-clockwise from the positive x-axis, the x-coordinate (which represents the cosine value) remains the same. So, $\cos(-\frac{\pi}{16})$ is indeed equal to $\cos(\frac{\pi}{16})$. Never let a negative sign inside a cosine function intimidate you!

Q: Why can't we just use the half-angle formula $\sin^2 \theta = \frac{1-\cos 2\theta}{2}$?

You certainly could! That is a perfectly valid path. However, using that formula would introduce multiple cosine terms with different arguments, which might make the algebra more cumbersome. The identity $\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B)$ is a more direct 'shortcut' that collapses the expression into a product, which is often much easier to handle in trigonometry problems.

Question

If $L = \sin^2 (\frac{\pi}{16}) - \sin^2 (\frac{\pi}{8})$ and $M = \cos^2 (\frac{\pi}{16}) - \sin^2 (\frac{\pi}{8})$, then

Options:

(A) $L = -\frac{1}{2\sqrt{2}} + \frac{1}{2} \cos \frac{\pi}{8}$

(B) $M = \frac{1}{2\sqrt{2}} + \frac{1}{2} \cos \frac{\pi}{8}$

(C) $M = \frac{1}{4\sqrt{2}} + \frac{1}{4} \cos \frac{\pi}{8}$

(D) $L = \frac{1}{4\sqrt{2}} - \frac{1}{4} \cos \frac{\pi}{8}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a vast, complex trigonometric landscape. You see two expressions, $L$ and $M$, filled with squares of sine and cosine at angles like $\frac{\pi}{16}$ and $\frac{\pi}{8}$.

At first glance, it feels like a chaotic mess of numbers. But as an elite JEE aspirant, you know that beneath this complexity lies a hidden, elegant structure waiting to be revealed.

Let us embark on this journey to simplify:
$$M = \cos^2 \left( \frac{\pi}{16} \right) - \sin^2 \left( \frac{\pi}{8} \right)$$

### The Anatomy of the Expression

Why do we start with $M$? Because in the world of competitive exams, we must be tactical. The options provided are heavily focused on $M$.

When we look at $M = \cos^2 A - \sin^2 B$, we see a familiar pattern. This is not just a random collection of terms; it is a classic trigonometric identity waiting to be unleashed.

The identity:
$$\cos^2 A - \sin^2 B = \cos(A+B) \cos(A-B)$$
This is one of the most powerful tools in your arsenal. It allows us to convert a difference of squares into a product, which is almost always easier to simplify.

### The Power of Transformation

Let us set $A = \frac{\pi}{16}$ and $B = \frac{\pi}{8}$. Substituting these into our identity, we get:
$$M = \cos\left( \frac{\pi}{16} + \frac{\pi}{8} \right) \cos\left( \frac{\pi}{16} - \frac{\pi}{8} \right)$$

Now, let us handle the angles. For the sum, we need a common denominator of $16$, so $\frac{\pi}{16} + \frac{2\pi}{16} = \frac{3\pi}{16}$.

For the difference, we have $\frac{\pi}{16} - \frac{2\pi}{16} = -\frac{\pi}{16}$.

Here is where your conceptual clarity shines: since cosine is an even function, $\cos(-\frac{\pi}{16}) = \cos(\frac{\pi}{16})$. The negative sign simply vanishes, leaving us with:
$$M = \cos\left( \frac{3\pi}{16} \right) \cos\left( \frac{\pi}{16} \right)$$

### The Final Polish

We are almost there! We have a product of two cosines, but the angles are not standard. To resolve this, we use the product-to-sum identity.

We multiply and divide by $2$ to get:
$$M = \frac{1}{2} \left[ 2 \cos\left( \frac{3\pi}{16} \right) \cos\left( \frac{\pi}{16} \right) \right]$$

Applying the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$, we get:
$$M = \frac{1}{2} \left[ \cos\left( \frac{3\pi}{16} + \frac{\pi}{16} \right) + \cos\left( \frac{3\pi}{16} - \frac{\pi}{16} \right) \right]$$

Simplifying the angles, we get $\frac{4\pi}{16} = \frac{\pi}{4}$ and $\frac{2\pi}{16} = \frac{\pi}{8}$. Thus:
$$M = \frac{1}{2} \left[ \cos\left( \frac{\pi}{4} \right) + \cos\left( \frac{\pi}{8} \right) \right]$$

Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, our final expression becomes:
$$M = \frac{1}{2} \left[ \frac{1}{\sqrt{2}} + \cos\left( \frac{\pi}{8} \right) \right]$$

This simplifies to the final result:
**$$M = \frac{1}{2\sqrt{2}} + \frac{1}{2} \cos\left( \frac{\pi}{8} \right)$$**

This matches one of our options perfectly. You have successfully navigated the complexity and emerged victorious!

Animated Solution for Mathematics - Trigonometry: If $L = sin^{2} (\frac{π}{16}) - sin^{2} (\frac{π}{8})$ and $M = cos^{2} (\frac{π}{16}) - sin^{2} (\frac{π}{8})$ , then

Select Answer:

Visualized Solution

Problem Statement

Analyzing Expression $M$

Applying the Trigonometric Identity

Simplifying the First Angle

Simplifying the Second Angle

The Product Expression for $M$

Product to Sum Transformation

Simplifying the New Angles

Substituting Known Values

Final Calculation for $M$

Conclusion and Final Answer

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Anatomy of the Expression

The Power of Transformation

The Final Polish

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If L=sin2(16π​)−sin2(8π​) and M=cos2(16π​)−sin2(8π​), then