The value of cos² 10° - cos 10° cos 50° + cos² 50° is (1) 3/2(1 + cos 20°) (2) 3/4 (3) 3/4 + cos 20°

Q: Wait, why do we use power reduction? Can't we just use $\cos(A-B)$ or something else?

Great question! While there are many paths in trigonometry, power reduction is the most efficient way to handle squared terms when the angles aren't standard. It transforms the degree of the expression from 2 to 1, making it linear and much easier to manipulate. Think of it as simplifying the complexity of the expression before you even start adding terms.

Q: How do I know which identity to use when I see a mess of cosines?

Pattern recognition is the heart of JEE trigonometry. If you see $\cos^2 \theta$, think Power Reduction. If you see $\cos A \cos B$, think Product-to-Sum. If you see a sum of cosines, think Sum-to-Product. It's like having a toolbox—you just need to identify the right tool for the specific shape of the problem.

Q: Why does the whole trigonometric part vanish at the end? Is that a coincidence?

In JEE Advanced problems, it is rarely a coincidence! When you see a complex expression simplifying to a clean constant like $3/4$, it's a sign that the trigonometric terms were designed to cancel out. Trust the process, follow the identities, and the 'magic' of the cancellation will reveal itself.

Question

The value of $\cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ$ is

Options:

(A) $\frac{3}{2}(1 + \cos 20^\circ)$

(B) $3/4$

(C) $\frac{3}{4} + \cos 20^\circ$

(D) $3/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow JEE warrior! Today, we are going to tackle a problem that might look like a chaotic mess of angles at first glance, but beneath the surface, it is a masterclass in symmetry and identity application.

We are looking at the expression:
$$\cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ$$

When you see angles like $10^\circ$ and $50^\circ$, your first instinct might be to reach for a calculator, but in the JEE exam hall, your brain is the only tool you have. Let's break this down.

### Phase 1: The Power Reduction Strategy

The expression contains squared terms, $\cos^2 10^\circ$ and $\cos^2 50^\circ$. These are the 'monsters' of the expression, and we need to tame them.

The most powerful weapon in our arsenal for squared trigonometric functions is the Power Reduction Identity:
$$\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$$

By applying this, we transform the squared terms into linear terms with double the angle. For the first term, $\cos^2 10^\circ$, we get:
$$\frac{1 + \cos 20^\circ}{2}$$

For the last term, $\cos^2 50^\circ$, we get:
$$\frac{1 + \cos 100^\circ}{2}$$

Suddenly, the expression is no longer about squares; it is about linear cosines, which are much easier to handle.

### Phase 2: The Product-to-Sum Bridge

Now, let's look at the middle term: $-\cos 10^\circ \cos 50^\circ$. This is a product of two cosines, and we need to break this product into a sum.

We use the Product-to-Sum identity:
$$\cos A \cos B = \frac{1}{2}(\cos(A+B) + \cos(A-B))$$

Setting $A = 10^\circ$ and $B = 50^\circ$, we get:
$$\frac{1}{2}(\cos 60^\circ + \cos(-40^\circ))$$

Since $\cos(-	heta) = \cos 	heta$, this simplifies to $\frac{1}{2}(\cos 60^\circ + \cos 40^\circ)$. Given that $\cos 60^\circ = \frac{1}{2}$, the term becomes:
$$\frac{1}{2}\left(\frac{1}{2} + \cos 40^\circight) = \frac{1}{4} + \frac{1}{2}\cos 40^\circ$$

### Phase 3: The Grand Cancellation

Now, let's assemble our pieces. The expression is now:
$$\frac{1 + \cos 20^\circ}{2} - \left(\frac{1}{4} + \frac{1}{2}\cos 40^\circight) + \frac{1 + \cos 100^\circ}{2}$$

Let's group the constants:
$$\frac{1}{2} - \frac{1}{4} + \frac{1}{2} = 1 - \frac{1}{4} = \frac{3}{4}$$

Now for the trigonometric part:
$$\frac{1}{2}\cos 20^\circ - \frac{1}{2}\cos 40^\circ + \frac{1}{2}\cos 100^\circ$$

Factoring out $\frac{1}{2}$, we have $\frac{1}{2}(\cos 20^\circ + \cos 100^\circ - \cos 40^\circ)$.

### Final Calculation

Here is the final act of magic. We use the Sum-to-Product identity on $\cos 100^\circ + \cos 20^\circ$:
$$2\cos\left(\frac{100^\circ+20^\circ}{2}ight)\cos\left(\frac{100^\circ-20^\circ}{2}ight) = 2\cos 60^\circ \cos 40^\circ$$

Since $\cos 60^\circ = \frac{1}{2}$, this simplifies to $2 	imes \frac{1}{2} 	imes \cos 40^\circ = \cos 40^\circ$. Substituting this back, we get:
$$\frac{1}{2}(\cos 40^\circ - \cos 40^\circ) = 0$$

The entire trigonometric part vanishes! We are left with just **$\frac{3}{4}$**. This is the elegance of mathematics—a complex expression collapsing into a simple, beautiful constant.

Animated Solution for Mathematics - Trigonometry: The value of $cos^{2} 1 0^{\circ} - cos 1 0^{\circ} cos 5 0^{\circ} + cos^{2} 5 0^{\circ}$ is

Select Answer:

Visualized Solution

The Given Expression

The Mathematical Toolkit

Transforming the First Term

Transforming the Middle Term

Transforming the Last Term

Combining All Terms

Separating Constants and Trig

Simplifying the Constants

Applying Sum-to-Product

The Final Cancellation

Conclusion and Final Answer

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

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