How do you simplify the terms sin(7π/2 - θ) and sin(11π + θ)?

Using quadrant rules, sin(7π/2 - θ) = -cos θ and sin(11π + θ) = -sin θ. Since the powers are even, they become cos⁴ θ and sin⁴ θ respectively.

What are the simplified expressions for the sin⁶ terms?

Similarly, sin(3π/2 - θ) = -cos θ and sin(9π - θ) = sin θ. Raising these to the 6th power results in cos⁶ θ and sin⁶ θ.

What is the simplified form of the function f(θ)?

The function simplifies to f(θ) = 4(sin⁴ θ + cos⁴ θ) - 2(sin⁶ θ + cos⁶ θ).

Which identity is used for sin⁴ θ + cos⁴ θ?

We use the identity sin⁴ θ + cos⁴ θ = 1 - 2 sin² θ cos² θ.

Which identity is used for sin⁶ θ + cos⁶ θ?

We use the identity sin⁶ θ + cos⁶ θ = 1 - 3 sin² θ cos² θ.

Is the function f(θ) actually a constant?

Yes, after substituting the identities, all terms involving θ cancel out, leaving f(θ) = 2.

What are the values of α and β?

Since f(θ) = 2 for all real θ, the maximum value α is 2 and the minimum value β is also 2.

How do we find α + 2β?

Substituting α = 2 and β = 2 into the expression, we get 2 + 2(2) = 6.

Why is the answer 6 and not 5 as shown in some keys?

Rigorous simplification shows the function is constant at 2. Therefore, α + 2β = 2 + 4 = 6. Always trust the mathematical reduction.

Is this a common pattern in JEE Main?

Yes, problems that appear complex but simplify to constants using identities are very common in trigonometry and calculus sections.

Let α and β respectively be the maximum and the minimum values of the function f(θ) = 4 (sin⁴ (7π/2 - θ) + sin⁴ (11π + θ)) - 2 (sin⁶ (3π/2 - θ) + sin⁶ (9π

DOCTYPE html> Let α and β respectively be the maximum and the minimum values of the function f(θ) = 4 (sin⁴ (7π/2 – θ) + sin⁴ (11π + θ)) – 2 (sin⁶ (3π/2 – θ) + sin⁶ (9π – θ)), θ ∈ R. Then α + 2β is equal to | JEE Main Mathematics

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Math Trigonometry

Q MCQ Trig Identities

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

A) $6$ B) $5$ C) $4$ D) $3$

✅ Correct Answer

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Solution Steps

Simplify individual trigonometric terms

Using reduction formulas for each term:

$\sin(\frac{7\pi}{2} – \theta) = \sin(4\pi – \frac{\pi}{2} – \theta) = \sin(-(\frac{\pi}{2} + \theta)) = -\cos \theta$

$\sin(11\pi + \theta) = -\sin \theta$

$\sin(\frac{3\pi}{2} – \theta) = -\cos \theta$

$\sin(9\pi – \theta) = \sin \theta$

Re-evaluate the function $f(\theta)$

Substitute these into the powers (even powers remove the negative signs):

$f(\theta) = 4(\cos^4 \theta + \sin^4 \theta) – 2(\cos^6 \theta + \sin^6 \theta)$

Apply Standard Identities

We know the following identities:

$\sin^4 \theta + \cos^4 \theta = 1 – 2\sin^2 \theta \cos^2 \theta$

$\sin^6 \theta + \cos^6 \theta = 1 – 3\sin^2 \theta \cos^2 \theta$

Substitute and Simplify

$f(\theta) = 4(1 – 2\sin^2 \theta \cos^2 \theta) – 2(1 – 3\sin^2 \theta \cos^2 \theta)$

$f(\theta) = 4 – 8\sin^2 \theta \cos^2 \theta – 2 + 6\sin^2 \theta \cos^2 \theta$

$f(\theta) = 2 – 2\sin^2 \theta \cos^2 \theta$

Actually, there is a known identity $3(\sin^4 \theta + \cos^4 \theta) – 2(\sin^6 \theta + \cos^6 \theta) = 1$. Let’s re-verify the coefficients.

$f(\theta) = 4(1 – 2s^2 c^2) – 2(1 – 3s^2 c^2) = 4 – 8s^2 c^2 – 2 + 6s^2 c^2 = 2 – 2s^2 c^2 = 2 – \frac{1}{2}\sin^2 2\theta$.

Find Max ($\alpha$) and Min ($\beta$)

Since $0 \le \sin^2 2\theta \le 1$:

$\alpha (\text{Max}) = 2 – \frac{1}{2}(0) = 2$

$\beta (\text{Min}) = 2 – \frac{1}{2}(1) = 1.5 = 3/2$

Calculate Final Value

$\alpha + 2\beta = 2 + 2(3/2) = 2 + 3 = 5$.

$\alpha + 2\beta = 5$ (Option B)

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Theory

1. Reduction Formulas in Trigonometry

Trigonometric functions of angles in the form $(n\pi/2 \pm \theta)$ can be simplified using quadrant rules. If $n$ is odd, the function changes to its co-function (sin becomes cos). If $n$ is even, the function remains the same. The sign is determined by the quadrant in which the angle lies. For even powers, as seen in this problem ($^4$ and $^6$), the sign of the original function is irrelevant as it always results in a positive value.

2. Higher Order Sine-Cosine Identities

The expressions $\sin^4 \theta + \cos^4 \theta$ and $\sin^6 \theta + \cos^6 \theta$ are frequently occurring patterns in JEE. They are simplified using the basic identity $\sin^2 \theta + \cos^2 \theta = 1$. Specifically, squaring this gives $(\sin^2 \theta + \cos^2 \theta)^2 = \sin^4 \theta + \cos^4 \theta + 2\sin^2 \theta \cos^2 \theta = 1$. Cubing it gives $(\sin^2 \theta + \cos^2 \theta)^3 = \sin^6 \theta + \cos^6 \theta + 3\sin^2 \theta \cos^2 \theta(\sin^2 \theta + \cos^2 \theta) = 1$.

3. Range of Sinusoidal Functions

The range of $\sin \theta$ is $[-1, 1]$, and consequently, the range of $\sin^2 \theta$ is $[0, 1]$. In problems involving $f(\theta) = A + B\sin^2(k\theta)$, the maximum and minimum values are found by substituting the extreme values of the squared term. If $B$ is negative, the maximum occurs when $\sin^2(k\theta) = 0$ and the minimum occurs when $\sin^2(k\theta) = 1$.

4. Transformation to Double Angles

The term $\sin^2 \theta \cos^2 \theta$ can be simplified using the double angle formula $\sin 2\theta = 2 \sin \theta \cos \theta$. Squaring this gives $\sin^2 2\theta = 4 \sin^2 \theta \cos^2 \theta$, so $\sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2 2\theta$. This reduction is essential for determining the range of complex trigonometric expressions because it reduces the number of variables to a single periodic function.

5. Maximum and Minimum of Functions

In calculus and trigonometry, finding $\alpha$ (max) and $\beta$ (min) often involves analyzing the function’s derivative or its simplified structure. For simple periodic forms, inspection of the range is sufficient. If the function is constant, $\alpha = \beta$. If the function depends on a square, the endpoints of the range of that square (usually 0 and 1) define the extrema.

6. Linear Combinations of Extrema

Competitive exams like JEE Main often ask for values like $\alpha + 2\beta$ or $2\alpha – \beta$ to test careful calculation beyond just finding the range. This requires the student to correctly identify which value corresponds to the maximum and which to the minimum, then apply the arithmetic weights specified in the query.

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FAQs

Is $f(\theta)$ always a constant in these types of questions?

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Not always. While some identities result in a constant (like $3(\sin^4\theta + \cos^4\theta) – 2(\sin^6\theta + \cos^6\theta) = 1$), changing the coefficients can make it variable, as seen here.

Why did $\sin(7\pi/2 – \theta)$ become $-\cos \theta$?

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$7\pi/2$ is an odd multiple of $\pi/2$, which changes sin to cos. $7\pi/2 – \theta$ lies in the 3rd quadrant where sine is negative.

Does the choice of $\theta$ affect the max and min?

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No, since $\theta \in \mathbf{R}$, the term $\sin^2 2\theta$ will eventually cover its entire range [0, 1].

What if the powers were odd, like $\sin^3 \theta$?

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The signs would matter significantly. Even powers are simpler because they eliminate the need to track quadrant-based negative signs.

How do I remember the $\sin^6 \theta + \cos^6 \theta$ identity?

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Think of it as $(a+b)(a^2 + b^2 – ab)$ where $a = \sin^2 \theta$ and $b = \cos^2 \theta$. Since $a+b=1$, it becomes $(\sin^2\theta + \cos^2\theta)^2 – 3\sin^2\theta\cos^2\theta$.

What is the value of $\alpha$ in this problem?

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The maximum value $\alpha$ is 2, occurring when $\sin 2\theta = 0$.

What is the value of $\beta$ in this problem?

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The minimum value $\beta$ is 1.5 (or 3/2), occurring when $\sin^2 2\theta = 1$.

Can I use $t = \sin^2 \theta$ to solve this?

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Yes, you can substitute $\cos^2 \theta = 1-t$ and find the range of the resulting quadratic in $t$ for $t \in [0, 1]$.

Why did the image show “5” as the answer?

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Because based on our simplified function $f(\theta) = 2 – \frac{1}{2}\sin^2 2\theta$, $\alpha=2$ and $\beta=1.5$, thus $\alpha+2\beta = 2+3=5$.

Is this from 2026 JEE paper?

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The screenshot indicates it is from the 23rd January 2026 Morning Shift.

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