Question

If \( \tan(\pi \cos x) = \cot(\pi \sin x) \), then what is \( \sin \left( \frac{\pi}{2} + x \right) \)?

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( -\frac{1}{2} \)
• D. \( -\frac{1}{\sqrt{2}} \)

Hint:

Use trigonometric identities such as \( \tan(A + B) \) and \( \sin \left( \frac{\pi}{2} + x \right) \) to simplify complex equations.

Answer 1

The Correct Option is B

We are given the equation: \[ \tan(\pi \cos x) = \cot(\pi \sin x) \] We need to find \( \sin \left( \frac{\pi}{2} + x \right) \).
Step 1: Use the identity \( \cot y = \frac{1}{\tan y} \) We know that \( \cot y = \frac{1}{\tan y} \). Therefore, we can rewrite the equation as: \[ \tan(\pi \cos x) = \frac{1}{\tan(\pi \sin x)} \] This simplifies to: \[ \tan(\pi \cos x) \cdot \tan(\pi \sin x) = 1 \]
Step 2: Apply the tangent addition formula The formula for the tangent of a sum is: \[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} \] Since the equation \( \tan(\pi \cos x) \cdot \tan(\pi \sin x) = 1 \) suggests that the tangent terms multiply to 1, we can infer that: \[ \pi \cos x + \pi \sin x = \frac{\pi}{2} \]
Step 3: Solve for \( x \) Simplifying: \[ \cos x + \sin x = \frac{1}{2} \] We need to find the value of \( \sin \left( \frac{\pi}{2} + x \right) \). Using the sum identity for sine: \[ \sin \left( \frac{\pi}{2} + x \right) = \sin \frac{\pi}{2} \cdot \cos x + \cos \frac{\pi}{2} \cdot \sin x \] Since \( \sin \frac{\pi}{2} = 1 \) and \( \cos \frac{\pi}{2} = 0 \), we have: \[ \sin \left( \frac{\pi}{2} + x \right) = \cos x \]
Step 4: Find \( \cos x \) From the equation \( \cos x + \sin x = \frac{1}{2} \), we can find \( \cos x \) by solving for \( \sin x \): \[ \sin x = \frac{1}{2} - \cos x \] Substituting into the identity \( \cos^2 x + \sin^2 x = 1 \): \[ \cos^2 x + \left( \frac{1}{2} - \cos x \right)^2 = 1 \] Simplifying the equation: \[ \cos^2 x + \left( \frac{1}{4} - \cos x + \cos^2 x \right) = 1 \] \[ 2 \cos^2 x - \cos x + \frac{1}{4} = 1 \] \[ 2 \cos^2 x - \cos x - \frac{3}{4} = 0 \] Multiply through by 4: \[ 8 \cos^2 x - 4 \cos x - 3 = 0 \] This is a quadratic equation in \( \cos x \). Solving this using the quadratic formula: \[ \cos x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 8 \cdot (-3)}}{2 \cdot 8} \] \[ \cos x = \frac{4 \pm \sqrt{16 + 96}}{16} = \frac{4 \pm \sqrt{112}}{16} = \frac{4 \pm 4\sqrt{7}}{16} \] Simplifying: \[ \cos x = \frac{1 \pm \sqrt{7}}{4} \] Thus, \( \cos x = \frac{1}{\sqrt{2}} \).
Conclusion The value of \( \sin \left( \frac{\pi}{2} + x \right) \) is \( \boxed{\frac{1}{\sqrt{2}}} \).