But earlier we said $ f(x) = 1 - rac{1}{2}(\cos 2x - \cos 4x) $. When $ \cos 2x = -1 $, $ \cos 4x = 1 $, so $ -1 - 1 = -2 $, multiplied by $ -rac{1}{2} $ gives $ +1 $, so $ 1 + 1 = 2 $. Yes.

["Understanding the Identity: A Closer Look at $ f(x) = 1 - \frac{1}{2}(\cos 2x - \cos 4x) $", "Mathematics often reveals elegant identities that simplify complex expressions, and one such fascinating function is:", "$$\nf(x) = 1 - \frac{1}{2}(\cos 2x - \cos 4x)\n$$", "At first glance, this formula may appear abstract, but deeper analysis reveals a powerful identity that simplifies neatly. Let’s unpack it step by step — particularly exploring the case when $ \cos 2x = -1 $, where $ \cos 4x = 1 $ — and explain why $ f(x) = 2 $ under this condition.", "---", "### Step 1: Simplify the Expression Using Trigonometric Identities", "The function contains $ \cos 2x $ and $ \cos 4x $, which are double-angle related. Recall the double-angle identity for cosine:", "$$\n\cos 4x = 2\cos^2 2x - 1\n$$", "This helps eliminate ambiguity in values. Let’s substitute into $ f(x) $:", "$$\nf(x) = 1 - \frac{1}{2}\left(\cos 2x - (2\cos^2 2x - 1)\right)\n= 1 - \frac{1}{2}\left( \cos 2x - 2\cos^2 2x + 1 \right)\n$$", "Simplify inside the parentheses:", "$$\nf(x) = 1 - \frac{1}{2}\left( -2\cos^2 2x + \cos 2x + 1 \right)\n$$", "Distribute the $ -\frac{1}{2} $:", "$$\nf(x) = 1 + \cos^2 2x - \frac{1}{2}\cos 2x - \frac{1}{2}\n$$", "Combine constants:", "$$\nf(x) = \cos^2 2x - \frac{1}{2}\cos 2x + \frac{1}{2}\n$$", "This quadratic form in $ \cos 2x $ confirms $ f(x) $ is a smoothly varying function depending solely on $ \cos 2x $.", "---", "### Step 2: Evaluate $ f(x) $ When $ \cos 2x = -1 $", "Now consider the specific case $ \cos 2x = -1 $. Then from double-angle identities:", "$$\n\cos 4x = 2(-1)^2 - 1 = 2(1) - 1 = 1\n$$", "Plug into the original definition:", "$$\nf(x) = 1 - \frac{1}{2}(-1 - 1) = 1 - \frac{1}{2}(-2) = 1 + 1 = 2\n$$", "This confirms the statement: when $ \cos 2x = -1 $, $ f(x) = 2 $.", "---", "### Step 3: Explore the Broader Identity", "Let’s verify the earlier causal chain:\nIf $ \cos 2x = -1 $, then:", "$$\n\cos 4x = 1\n\Rightarrow \cos 2x - \cos 4x = -1 - 1 = -2\n\Rightarrow -\frac{1}{2}(\cos 2x - \cos 4x) = -\frac{1}{2}(-2) = +1\n\Rightarrow f(x) = 1 + 1 = 2\n$$", "This confirms consistency. Moreover, using the simplified polynomial form:", "$$\nf(x) = (-1)^2 - \frac{1}{2}(-1) + \frac{1}{2} = 1 + \frac{1}{2} + \frac{1}{2} = 2\n$$", "Hence, the identity holds exactly at this point.", "---", "### Why This Identity Matters", "Expressions like $ f(x) $ arise frequently in Fourier analysis, signal processing, and solving differential equations. Recognizing that $ f(x) $ evaluates to 2 whenever $ \cos 2x = -1 $ lets mathematicians and physicists leverage symmetry and periodicity more effectively.", "Such evaluations also help identify pivotal points — critical values where functions attain extreme or constant behavior — aiding in graphing and optimization.", "---", "### Conclusion", "The function\n$$\nf(x) = 1 - \frac{1}{2}(\cos 2x - \cos 4x)\n$$\ndemonstrates a precise and elegant numerical identity. When $ \cos 2x = -1 $, thanks to the trigonometric coupling and algebraic manipulation, it reliably yields $ f(x) = 2 $. Understanding these relationships reveals not just a formula, but a window into deeper mathematical patterns.", "Next time you encounter $ f(x) $, recall: at $ \cos 2x = -1 $, $ f(x) = 2 $—a small but powerful fact rooted in trigonometric identities.", "---", "Keywords: $ f(x) = 1 - \frac{1}{2}(\cos 2x - \cos 4x) $, identity evaluation, $ \cos 2x = -1 $, trigonometric simplification, Fourier identity, critical evaluation point, mathematical elegance."]