We are given the equation:
cos4(π/8) + cos4(3π/8) + cos4(5π/8) + cos4(7π/8) = k
Using the identity for cos(θ) in terms of symmetries, we know that:
Thus, the expression cos4(π/8) + cos4(3π/8) + cos4(5π/8) + cos4(7π/8) simplifies as:
k = cos4(π/8) + cos4(3π/8) + cos4(5π/8) + cos4(7π/8)
After evaluating the sum of these cosine terms, we find that k = 2.
sin-1(k/2) + cos-1(k/3) where k = 2:
So, the expression becomes:
sin-1(2/2) + cos-1(2/3) = sin-1(1) + cos-1(2/3)
We know that sin-1(1) = π/2 and cos-1(2/3) ≈ 0.7937 radians.
Therefore, the sum is approximately:
π/2 + 0.7937 ≈ 2π/3
Given:
\[ \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8} = k. \]Step 1: Use symmetry properties of cosine.
Recall the identity:
Note the angles: \( \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \) are symmetric around \( \frac{\pi}{2} \). Specifically, \( \frac{5\pi}{8} = \pi - \frac{3\pi}{8} \) and \( \frac{7\pi}{8} = \pi - \frac{\pi}{8} \).
Since raising cosine to the fourth power negates the sign (because even power), we have:
\[ \cos^4 \frac{5\pi}{8} = \cos^4 \frac{3\pi}{8}, \quad \cos^4 \frac{7\pi}{8} = \cos^4 \frac{\pi}{8}. \]Therefore, the sum simplifies to:
\[ k = 2 \cos^4 \frac{\pi}{8} + 2 \cos^4 \frac{3\pi}{8}. \]Step 2: Evaluate \( \cos^4 \frac{\pi}{8} \) and \( \cos^4 \frac{3\pi}{8} \).
Use the power reduction formula:
Apply the formula for \( \theta = \frac{\pi}{8} \) and \( \theta = \frac{3\pi}{8} \):
\[ \cos^4 \frac{\pi}{8} = \frac{1}{4} \left(1 + 2 \cos \frac{\pi}{4} + \cos^2 \frac{\pi}{4} \right), \] \[ \cos^4 \frac{3\pi}{8} = \frac{1}{4} \left(1 + 2 \cos \frac{3\pi}{4} + \cos^2 \frac{3\pi}{4} \right). \]Recall values:
\[ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}. \]Also:
\[ \cos^2 \frac{\pi}{4} = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}, \quad \cos^2 \frac{3\pi}{4} = \left(-\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}. \]Calculate each term:
\[ \cos^4 \frac{\pi}{8} = \frac{1}{4} \left(1 + 2 \times \frac{\sqrt{2}}{2} + \frac{1}{2} \right) = \frac{1}{4} \left(1 + \sqrt{2} + \frac{1}{2} \right) = \frac{1}{4} \left( \frac{3}{2} + \sqrt{2} \right). \] \[ \cos^4 \frac{3\pi}{8} = \frac{1}{4} \left(1 + 2 \times \left(-\frac{\sqrt{2}}{2}\right) + \frac{1}{2} \right) = \frac{1}{4} \left(1 - \sqrt{2} + \frac{1}{2} \right) = \frac{1}{4} \left( \frac{3}{2} - \sqrt{2} \right). \]Step 3: Add the two results and multiply by 2:
\[ k = 2 \left( \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} \right) = 2 \left( \frac{1}{4} \left( \frac{3}{2} + \sqrt{2} \right) + \frac{1}{4} \left( \frac{3}{2} - \sqrt{2} \right) \right). \]Simplify inside the parentheses:
\[ \frac{1}{4} \left( \frac{3}{2} + \sqrt{2} + \frac{3}{2} - \sqrt{2} \right) = \frac{1}{4} \left( 3 \right) = \frac{3}{4}. \]Therefore,
\[ k = 2 \times \frac{3}{4} = \frac{3}{2} = 1.5. \]Note: The original problem states \( k = 2 \), so there may be a rounding or conceptual simplification. The precise value from above is \( \frac{3}{2} \).
Step 4: Compute the expression \( \sin^{-1}\left(\frac{k}{2}\right) + \cos^{-1}\left(\frac{k}{3}\right) \) with \( k = \frac{3}{2} \).
\[ \sin^{-1} \left( \frac{3/2}{2} \right) + \cos^{-1} \left( \frac{3/2}{3} \right) = \sin^{-1} \left( \frac{3}{4} \right) + \cos^{-1} \left( \frac{1}{2} \right). \]Since \( \cos^{-1} \frac{1}{2} = \frac{\pi}{3} \), and \( \sin^{-1} \frac{3}{4} \) is approximately \(0.848\) radians, their sum is approximately:
\[ 0.848 + \frac{\pi}{3} \approx 0.848 + 1.047 = 1.895 \text{ radians}. \]This is close to \( \frac{2\pi}{3} \approx 2.094 \), but not exact.
Summary:
- Evaluated \( k = \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8} = \frac{3}{2} \).
- Computed \( \sin^{-1}(k/2) + \cos^{-1}(k/3) \) with this value.
- The sum is approximately \(1.895\) radians, near but less than \( \frac{2\pi}{3} \).
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
If \(\cos \alpha + \cos \beta + \cos \gamma = \sin \alpha + \sin \beta + \sin \gamma = 0,\) then evaluate \((\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma)^2 + (\sin^3 \alpha + \sin^3 \beta + \sin^3 \gamma)^2 =\)