Question

If $\tan (\pi \cos x)=\cot (\pi \sin x)$, then the absolute value of $4 \sqrt{2} \cos \left(x-\frac{\pi}{4}\right)$ is:

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Answer


Given: $$ \tan (\pi \cos x) = \cot (\pi \sin x) $$

Step-by-Step :

  1. Convert cotangent to tangent: $$ \tan (\pi \cos x) = \tan \left( \frac{\pi}{2} - \pi \sin x \right) $$

  2. Use the identity for tangent: $$ \tan (\pi \cos x) = \tan \left( \frac{\pi}{2} - \pi \sin x \right) $$ This implies: $$ \pi \cos x = n\pi + \frac{\pi}{2} - \pi \sin x $$ (where $ n $ is an integer).

  3. Rearrange the equation: $$ \cos x + \sin x = \frac{2n + 1}{2} $$

  4. Factor out $ \frac{1}{\sqrt{2}} $: $$ \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = \frac{2n + 1}{2 \sqrt{2}} $$

  5. Apply the angle addition formula: $$ \cos \left( x - \frac{\pi}{4} \right) = \pm \frac{1}{2 \sqrt{2}} $$ (considering the range $ \cos x \in [-1, 1] $).

Therefore, the absolute value of: $$ 4 \sqrt{2} \cos \left( x - \frac{\pi}{4} \right) = 4 \sqrt{2} \left( \pm \frac{1}{2 \sqrt{2}} \right) = \pm 2 $$

Hence, the final answer is: $$ \boxed{2} $$


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