Let f : (-1, 1) → R be such that f(cos 4θ) = 2 / (2 - sec² θ) for θ ∈ (0, π/4) ∪ (π/4, π/2). Then th

Q: Wait, why do we reject the negative root here?

Actually, we don't! That's the trap. The domain of $\theta$ is $(0, \pi/4) \cup (\pi/4, \pi/2)$. This means $2\theta$ spans $(0, \pi/2) \cup (\pi/2, \pi)$. In this range, $\cos 2\theta$ can be both positive and negative. Always check the domain!

Q: Why can't we just use energy conservation?

This is a trigonometry problem, not physics! But the spirit is the same: look for the conservation of simplicity. Don't overcomplicate the algebra when a simple identity can solve it.

Q: Is there a faster way to solve this without simplifying the function first?

You could try, but you'd be dealing with $\sec^2 \theta$ and $\cos 4\theta$ simultaneously, which is a nightmare. Simplifying first is the 'golden path' in JEE problems.

Question

Let $f : (-1, 1) 	o R$ be such that $f(\cos 4	heta) = \frac{2}{2 - \sec^2 	heta}$ for $	heta \in (0, \pi/4) \cup (\pi/4, \pi/2)$. Then the value(s) of $f(1/3)$ is (are)

Options:

(A) $1 - \sqrt{3/2}$

(B) $1 + \sqrt{3/2}$

(C) $1 - \sqrt{2/3}$

(D) $1 + \sqrt{2/3}$

Sigma Percentile · Accepted Answer

### The Art of Trigonometric Simplification

Welcome, future engineer! Today, we are going to dismantle a trigonometric beast. Often in JEE Advanced, you will encounter problems that look like a tangled mess of secants, cosines, and double angles.

The secret to conquering them is not brute force, but the art of simplification. Let us look at our function: 
$$f(\cos 4	heta) = \frac{2}{2 - \sec^2 	heta}$$

### Phase 1: Clearing the Fog

At first glance, the presence of $\sec^2 	heta$ in the denominator is unsettling. But remember, $\sec 	heta$ is just the reciprocal of $\cos 	heta$. So, $\sec^2 	heta = \frac{1}{\cos^2 	heta}$.

Let us substitute this into our function:

$$f(\cos 4	heta) = \frac{2}{2 - \frac{1}{\cos^2 	heta}}$$

This is a complex fraction. To clean it up, we multiply both the numerator and the denominator by $\cos^2 	heta$. This is a classic move to eliminate the inner fraction:

$$f(\cos 4	heta) = \frac{2 \cos^2 	heta}{2 \cos^2 	heta - 1}$$

Now, look at that denominator: $2 \cos^2 	heta - 1$. Does it look familiar? It is the fundamental double-angle identity for cosine: $\cos 2	heta = 2 \cos^2 	heta - 1$.

Similarly, the numerator $2 \cos^2 	heta$ can be written as $1 + \cos 2	heta$. Substituting these, we get:

$$f(\cos 4	heta) = \frac{1 + \cos 2	heta}{\cos 2	heta} = \frac{1}{\cos 2	heta} + 1$$

### Phase 2: The Logical Bridge

We have simplified the function to $f(\cos 4	heta) = \frac{1}{\cos 2	heta} + 1$. Now, the question asks for $f(1/3)$. This means we need to set the input $\cos 4	heta = 1/3$.

But our function is in terms of $\cos 2	heta$. We need a bridge between $\cos 4	heta$ and $\cos 2	heta$. Again, we turn to our trusty double-angle identity:

$$\cos 4	heta = 2 \cos^2 2	heta - 1$$

Setting this equal to $1/3$ gives us:

$$2 \cos^2 2	heta - 1 = \frac{1}{3}$$

### Phase 3: The Final Calculation

Solving for $\cos^2 2	heta$ is straightforward algebra:

$$2 \cos^2 2	heta = 1 + \frac{1}{3} = \frac{4}{3}$$

$$\cos^2 2	heta = \frac{2}{3}$$

Taking the square root, we get $\cos 2	heta = \pm \sqrt{\frac{2}{3}}$. Because the domain of $	heta$ allows for both positive and negative values of $\cos 2	heta$, we must consider both cases.

For the positive root:
$$f(1/3) = 1 + \frac{1}{\sqrt{2/3}} = 1 + \sqrt{\frac{3}{2}}$$

For the negative root:
$$f(1/3) = 1 + \frac{1}{-\sqrt{2/3}} = 1 - \sqrt{\frac{3}{2}}$$

And there you have it! By simplifying the expression first, we turned a terrifying problem into a simple algebraic exercise. Keep this mindset, and you will dominate the math section!

Animated Solution for Mathematics - Trigonometry: Let $f : (- 1, 1) \to R$ be such that $f (cos 4 θ) = \frac{2}{2 - s e c ^{2} θ}$ for $θ \in (0, π /4) \cup (π /4, π /2)$ . Then the value(s) of $f (1/3)$ is (are)

Select Answer:

Visualized Solution

Analyzing the Given Function

Simplifying the Denominator

Clearing the Complex Fraction

Using $cos 2 θ$ Identity

Rewriting the Numerator

Separating the Terms

Equating the Input to $\frac{1}{3}$

Relating $cos 4 θ$ to $cos 2 θ$

Algebraic Manipulation

Taking the Square Root

Case 1: Positive Root

Case 2: Negative Root

The Final Answer

The Sigma Insight: Multiple and Sub-multiple Angles

The Art of Trigonometric Simplification

Phase 1

Phase 2

Phase 3

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