Let S = { θ ∈ [0, 2π] : 8^(2sin²θ) + 8^(2cos²θ) = 16 }. Then n(S) + Σ (sec(π/4 + 2θ) cosec(π/4 + 2θ)

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

Actually, we don't! When we solve $\sin^2 \theta = \frac{1}{2}$, we get $\sin \theta = \pm \frac{1}{\sqrt{2}}$. Both positive and negative values are valid because the square of both is $\frac{1}{2}$. This is why we have four solutions in the interval $[0, 2\pi]$.

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

This is a purely algebraic and trigonometric problem, not a physics problem involving energy. We rely on trigonometric identities and the properties of exponents to find the set of angles $S$.

Q: How do I handle the summation term?

The term $\sec(\frac{\pi}{4} + 2\theta) \csc(\frac{\pi}{4} + 2\theta)$ looks intimidating, but by converting it to $\frac{1}{\sin(\frac{\pi}{4} + 2\theta) \cos(\frac{\pi}{4} + 2\theta)}$ and multiplying the numerator and denominator by $2$, it simplifies to $\frac{2}{\sin(\frac{\pi}{2} + 4\theta)}$, which is just $\frac{2}{\cos 4\theta}$. It's all about simplification!

Question

Let $S = \{ 	heta \in [0, 2\pi] : 8^{2\sin^2 	heta} + 8^{2\cos^2 	heta} = 16 \}$. Then $n(S) + \sum_{	heta \in S} \left( \sec \left( \frac{\pi}{4} + 2	heta ight) \csc \left( \frac{\pi}{4} + 2	heta ight) ight)$ is equal to :

Options:

(A) 0

(B) -2

(C) -4

(D) 12

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given equation is $8^{2\sin^2 	heta} + 8^{2\cos^2 	heta} = 16$. To simplify this, we utilize the trigonometric identity $\cos^2 	heta = 1 - \sin^2 	heta$.

Substituting this into the equation, we obtain:

$$8^{2\sin^2 	heta} + 8^{2(1 - \sin^2 	heta)} = 16$$

This transformation allows us to express the entire equation in terms of a single trigonometric ratio.

### The Quadratic Reveal

Using the laws of indices, we rewrite the second term as $\frac{8^2}{8^{2\sin^2 	heta}}$. Let $y = 8^{2\sin^2 	heta}$. The equation transforms into:

$$y + \frac{64}{y} = 16$$

Multiplying the entire equation by $y$ yields the quadratic form $y^2 - 16y + 64 = 0$. This is a perfect square trinomial:

$$(y - 8)^2 = 0$$

Consequently, we find that $y = 8$.

### The Trigonometric Hunt

Returning to our original variable, we set $8^{2\sin^2 	heta} = 8^1$. Equating the exponents gives $2\sin^2 	heta = 1$, which simplifies to:

$$\sin^2 	heta = \frac{1}{2} \Rightarrow \sin 	heta = \pm \frac{1}{\sqrt{2}}$$

In the interval $[0, 2\pi]$, this condition is satisfied by the set of angles $S = \{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \}$. Thus, the number of elements is $n(S) = 4$.

### The Summation Symphony

We evaluate the expression $T = \sec(\frac{\pi}{4} + 2	heta) \csc(\frac{\pi}{4} + 2	heta)$. Converting to sine and cosine, we have:

$$T = \frac{1}{\sin(\frac{\pi}{4} + 2	heta) \cos(\frac{\pi}{4} + 2	heta)}$$

Multiplying the numerator and denominator by $2$ and applying the identity $\sin(2A) = 2\sin A \cos A$, we get:

$$T = \frac{2}{\sin(\frac{\pi}{2} + 4	heta)} = \frac{2}{\cos 4	heta}$$

For every $	heta \in S$, the value of $4	heta$ results in an odd multiple of $\pi$ (specifically $\pi, 3\pi, 5\pi, 7\pi$). Since $\cos(	ext{odd } \pi) = -1$, we find $T = -2$ for all four values.

### Final Calculation

The sum of the expression over all $	heta \in S$ is:

$$\sum T = 4 	imes (-2) = -8$$

Finally, combining this with the count of the set $S$:

$$n(S) + \sum T = 4 + (-8) = -4$$

The final answer is **-4**.

Animated Solution for Mathematics - Trigonometry: Let $S = {θ \in [0, 2 π] : 8^{2 s i n^{2} θ} + 8^{2 c o s^{2} θ} = 16}$ . Then $n (S) + \sum_{θ \in S} (sec (\frac{π}{4} + 2 θ) csc (\frac{π}{4} + 2 θ))$ is equal to :

Select Answer:

Visualized Solution

Analyze the Equation

Apply Trigonometric Identity

Simplify Exponents

Substitution for Quadratic Form

Solve for $y$

Back-Substitution

Find Values of $sin θ$

Identify the Set $S$

Analyze the Summation Term

Simplify Using Double Angle Formula

Final Form of the Term

Evaluate $cos 4 θ$

Calculate Individual Term Value

Final Calculation

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

The Quadratic Reveal

The Trigonometric Hunt

The Summation Symphony

Final Calculation

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={θ∈[0,2π]:82sin2θ+82cos2θ=16}. Then n(S)+∑θ∈S​(sec(4π​+2θ)csc(4π​+2θ)) is equal to :

Select Answer:

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Back-Substitution

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The Quadratic Reveal

The Trigonometric Hunt

The Summation Symphony

Final Calculation

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