Let α and β respectively be the maximum and the minimum values of the function f(θ) = 4 (sin⁴(7π/2 -

Q: Wait, why do we reject the negative signs when raising to the power of 4 or 6?

That is a great observation! Remember that any real number raised to an even power becomes positive. Since we are dealing with $\sin^4$ and $\sin^6$, the negative signs resulting from the reduction formulas (like $\sin(\frac{7\pi}{2} - \theta) = -\cos\theta$) are eliminated. $(-\cos\theta)^4 = \cos^4\theta$ and $(-\cos\theta)^6 = \cos^6\theta$. It is a classic trap to forget this, so always check the exponent!

Q: Why can't we just use energy conservation or differentiate the function?

While differentiation is a valid mathematical approach, it is often a 'brute force' method in JEE Advanced. Differentiating this expression would lead to a complex chain rule problem, increasing the probability of calculation errors. The 'JEE way' is to recognize the underlying algebraic structure—the sum of powers of sine and cosine—and use identities to simplify the expression into a single trigonometric ratio. It is faster, cleaner, and much less prone to error.

Q: How do I know which identity to use for $\sin^4\theta + \cos^4\theta$?

These identities are derived from the fundamental identity $\sin^2\theta + \cos^2\theta = 1$. If you square both sides, you get $(\sin^2\theta + \cos^2\theta)^2 = 1^2$, which expands to $\sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta = 1$. Rearranging this gives $\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$. Similarly, you can cube the fundamental identity to find the formula for the 6th powers. Memorizing these is a huge advantage in competitive exams.

Question

Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(	heta) = 4 \left(\sin^4 \left(\frac{7\pi}{2} - 	hetaight) + \sin^4(11\pi + 	heta)ight) - 2 \left(\sin^6 \left(\frac{3\pi}{2} - 	hetaight) + \sin^6(9\pi - 	heta)ight), 	heta \in \mathbb{R}$. Then $\alpha + 2\beta$ is equal to :

Options:

(A) 4

(B) 6

(C) 5

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The expression $f(	heta) = 4 (\sin^4 (\frac{7\pi}{2} - 	heta) + \sin^4(11\pi + 	heta)) - 2 (\sin^6 (\frac{3\pi}{2} - 	heta) + \sin^6(9\pi - 	heta))$ appears intimidating at first glance. In the context of JEE Advanced, such problems are designed to test your ability to simplify complex expressions through reduction rather than brute force.

### Phase 1: The Cleansing of Angles

Our first mission is to simplify the angles using reduction formulas. For $\sin(\frac{7\pi}{2} - 	heta)$, we note that $\frac{7\pi}{2} = 3\pi + \frac{\pi}{2}$, which places the angle in the third quadrant where sine is negative. The $\frac{\pi}{2}$ shift converts sine to cosine, yielding $\sin(\frac{7\pi}{2} - 	heta) = -\cos	heta$.

Raising this to the power of 4, the negative sign vanishes:
$$\sin^4\left(\frac{7\pi}{2} - 	hetaight) = \cos^4	heta$$

Similarly, $\sin(11\pi + 	heta)$ is in the third quadrant, becoming $-\sin	heta$. Raising this to the power of 4 gives $\sin^4	heta$.

For the 6th power terms, $\sin(\frac{3\pi}{2} - 	heta)$ becomes $-\cos	heta$ and $\sin(9\pi - 	heta)$ becomes $\sin	heta$. Raising these to the power of 6 yields $\cos^6	heta$ and $\sin^6	heta$. Our function now simplifies to:
$$f(	heta) = 4(\cos^4	heta + \sin^4	heta) - 2(\cos^6	heta + \sin^6	heta)$$

### Phase 2: The Algebraic Beauty

We avoid differentiation by utilizing standard algebraic identities. We know that:
$$\sin^4	heta + \cos^4	heta = 1 - 2\sin^2	heta\cos^2	heta$$
$$\sin^6	heta + \cos^6	heta = 1 - 3\sin^2	heta\cos^2	heta$$

Substituting these into our function:
$$f(	heta) = 4(1 - 2\sin^2	heta\cos^2	heta) - 2(1 - 3\sin^2	heta\cos^2	heta)$$

Expanding and combining the terms:
$$f(	heta) = 4 - 8\sin^2	heta\cos^2	heta - 2 + 6\sin^2	heta\cos^2	heta$$
$$f(	heta) = 2 - 2\sin^2	heta\cos^2	heta$$

### Phase 3: The Final Synthesis

To find the range, we express the function in terms of a single trigonometric ratio. Using the double-angle identity $\sin(2	heta) = 2\sin	heta\cos	heta$, we square both sides to get $\sin^2(2	heta) = 4\sin^2	heta\cos^2	heta$.

This allows us to rewrite the function as:
$$f(	heta) = 2 - \frac{1}{2}\sin^2(2	heta)$$

The maximum value $\alpha$ occurs when $\sin^2(2	heta) = 0$:
$$\alpha = 2 - 0 = 2$$

The minimum value $\beta$ occurs when $\sin^2(2	heta) = 1$:
$$\beta = 2 - \frac{1}{2} = \frac{3}{2}$$

Finally, we calculate $\alpha + 2\beta$:
$$\alpha + 2\beta = 2 + 2\left(\frac{3}{2}ight) = 2 + 3 = 5$$

The final answer is **5**.

Select Answer:

Visualized Solution

Analyze the Function $f (θ)$

Simplify $4^{th}$ Power Terms

Simplify $6^{th}$ Power Terms

Rewrite the Function $f (θ)$

Apply Algebraic Identities

Substitute and Expand

Simplify to a Single Variable

Find Maximum Value $α$

Find Minimum Value $β$

Calculate Final Answer $α + 2 β$

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Visualized Solution

Analyze the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(θ)

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4th Power Terms

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 6th Power Terms

Rewrite the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(θ)

Apply Algebraic Identities

Substitute and Expand

Simplify to a Single Variable

Find Maximum Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Find Minimum Value .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Calculate Final Answer .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α+2β