The least value of (cos² θ - 6 sin θ cos θ + 3 sin² θ + 2) is (1) -1 (2) 1 (3) 4 - √10 (4) 4 + √10

Q: Why can't I just differentiate this with respect to $\theta$ and set it to zero?

You absolutely can! Calculus is a powerful tool. However, differentiating $\cos^2 \theta - 6 \sin \theta \cos \theta + 3 \sin^2 \theta + 2$ involves the chain rule and product rule, leading to a more complex equation. The trigonometric identity approach is elegant and less prone to algebraic errors—it's about choosing the path of least resistance.

Q: How do I know the range of $a \cos x + b \sin x$ is $\pm \sqrt{a^2 + b^2}$?

Think of it as a vector projection. If you write $a \cos x + b \sin x$ as $\sqrt{a^2 + b^2} (\frac{a}{\sqrt{a^2 + b^2}} \cos x + \frac{b}{\sqrt{a^2 + b^2}} \sin x)$, you can treat $\frac{a}{\sqrt{a^2 + b^2}}$ as $\sin \phi$ and $\frac{b}{\sqrt{a^2 + b^2}}$ as $\cos \phi$. This becomes $\sqrt{a^2 + b^2} \sin(x + \phi)$, which oscillates between $\pm \sqrt{a^2 + b^2}$.

Q: Wait, why did we subtract the maximum value to find the minimum of the whole expression?

Look at the simplified expression: $E = 4 - (\cos 2\theta + 3 \sin 2\theta)$. Since we are subtracting a value from 4, to make $E$ as small as possible, we must subtract the *largest possible* value of the term in the parentheses. It's like having a bank account balance of 4 dollars and wanting to minimize it by withdrawing as much as possible!

Question

The least value of $(\cos^2 	heta - 6 \sin 	heta \cos 	heta + 3 \sin^2 	heta + 2)$ is

Options:

(A) -1

(B) 1

(C) 4 - \sqrt{10}

(D) 4 + \sqrt{10}

Sigma Percentile · Accepted Answer

### The Art of Trigonometric Simplification

Welcome, fellow traveler on the JEE journey. Today, we are going to dismantle a trigonometric expression that, at first glance, looks like a tangled mess of squares and products.

The expression is $E = \cos^2 	heta - 6 \sin 	heta \cos 	heta + 3 \sin^2 	heta + 2$. Our mission is to find its absolute minimum value.

Many students see this and immediately reach for the derivative, but I want to show you a more elegant, more powerful way to see the underlying geometry of this function.

### The Strategy: Linearization

When you encounter squared trigonometric terms like $\cos^2 	heta$ and $\sin^2 	heta$, they are often hiding the true nature of the function. They are essentially 'second-degree' terms that obscure the simple oscillation happening underneath.

The secret weapon here is the double-angle identity. By converting these squared terms into linear terms of $2	heta$, we can collapse the complexity.

We know that:
$$\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$$
$$\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$$
$$2 \sin 	heta \cos 	heta = \sin 2	heta$$

### The Execution: Collapsing the Expression

Let us perform the substitution with care. We take our expression $E$ and replace the terms:

$$E = \frac{1 + \cos 2	heta}{2} - 3(2 \sin 	heta \cos 	heta) + 3\left(\frac{1 - \cos 2	heta}{2}ight) + 2$$

Notice how we cleverly split the $-6 \sin 	heta \cos 	heta$ into $-3(2 \sin 	heta \cos 	heta)$ to perfectly match our identity. Now, let us expand this carefully:

$$E = \frac{1}{2} + \frac{1}{2}\cos 2	heta - 3 \sin 2	heta + \frac{3}{2} - \frac{3}{2}\cos 2	heta + 2$$

Take a breath here—don't rush the algebra. Group the constants: $\frac{1}{2} + \frac{3}{2} + 2 = 4$.

Now group the cosine terms: $\frac{1}{2}\cos 2	heta - \frac{3}{2}\cos 2	heta = -\cos 2	heta$. The expression has now simplified into something beautiful:

$$E = 4 - \cos 2	heta - 3 \sin 2	heta$$

### The Insight: Finding the Extremes

We can rewrite this as $E = 4 - (\cos 2	heta + 3 \sin 2	heta)$. To find the *least* value of $E$, we need to make the term being subtracted as large as possible.

This brings us to a fundamental JEE concept: the range of any expression of the form $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.

In our case, $a = 1$ and $b = 3$. The maximum value of $(\cos 2	heta + 3 \sin 2	heta)$ is:

$$\sqrt{1^2 + 3^2} = \sqrt{10}$$

Therefore, the minimum value of $E$ is **$4 - \sqrt{10}$**.

We have successfully navigated the complexity and arrived at the truth. Keep this strategy in your arsenal—whenever you see squares, think linear, and you will always find the path to the solution.

Animated Solution for Mathematics - Trigonometry: The least value of $(cos^{2} θ - 6 sin θ cos θ + 3 sin^{2} θ + 2)$ is

Select Answer:

Visualized Solution

The Expression $E$

Double Angle Strategy

Raw Substitution

Expanding the Terms

Grouping and Simplifying

Analyzing the Minimum

Range of $a cos x + b sin x$

Calculating the Maximum

The Final Minimum Value

The Way Forward

The Sigma Insight: Multiple and Sub-multiple Angles

The Art of Trigonometric Simplification

The Strategy

The Execution

The Insight

Similar Questions

If $sin θ + cos θ = \frac{1}{2}$ , then $16 (sin (2 θ) + cos (4 θ) + sin (6 θ))$ is equal to:

Value of $cos^{3} \frac{π}{8} cos \frac{3 π}{8} + sin^{3} \frac{π}{8} sin \frac{3 π}{8}$ is

The value of $cos^{3} \frac{π}{8} cos \frac{3 π}{8} + sin^{3} \frac{π}{8} sin \frac{3 π}{8}$ is:

The value of $cos \frac{π}{2 ^{2}} \cdot cos \frac{π}{2 ^{3}} \cdot ..... \cdot cos \frac{π}{2 ^{10}} \cdot sin \frac{π}{2 ^{10}}$ is :

Let $\frac{π}{2} < x < π$ be such that $cot x = - \frac{5}{11}$ . Then $sin (\frac{11 x}{2}) (sin 6 x - cos 6 x) + cos (\frac{11 x}{2}) (sin 6 x + cos 6 x)$ is equal to

The value of $36 (4 cos^{2} 9^{\circ} - 1) (4 cos^{2} 2 7^{\circ} - 1) (4 cos^{2} 8 1^{\circ} - 1) (4 cos^{2} 24 3^{\circ} - 1)$ is

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)

Let $α$ and $β$ be nonzero real numbers such that $2 (cos β - cos α) + cos α cos β = 1$ . Then which of the following is/are true?

Show that the value of $\frac{t a n x}{t a n 3 x}$ , wherever defined never lies between $1/3$ and 3.

Let $n$ be an odd integer. If $sin n θ = \sum_{r = 0}^{n} b_{r} sin^{r} θ$ , for every value of $θ$ , then

Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The least value of (cos2θ−6sinθcosθ+3sin2θ+2) is