\[ \sqrt{2} \sin(c) = -1 \quad \Rightarrow \quad \sin(c) = -\frac{1}{\sqrt{2}} \]

["Understanding the Equation ( \sqrt{2} \sin(c) = -1 ) and Solving for ( \sin(c) = -\frac{1}{\sqrt{2}} )", "When solving trigonometric equations, one of the most common transformations begins with isolating the sine function. Consider the equation:", "[\n\sqrt{2} \sin(c) = -1\n]", "To solve for ( \sin(c) ), divide both sides by ( \sqrt{2} ):", "[\n\sin(c) = -\frac{1}{\sqrt{2}}\n]", "This simplification is fundamental in trigonometry and enables us to analyze the behavior and solutions of the original equation more easily.", "---", "### Why Rewrite as ( \sin(c) = -\frac{1}{\sqrt{2}} )?", "Expressing the equation in terms of ( \sin(c) = -\frac{1}{\sqrt{2}} ) helps identify key properties of the sine function, including reference angles and corresponding standard angles. The value ( -\frac{1}{\sqrt{2}} ) appears frequently in the unit circle and is closely tied to standard angles like ( -\frac{\pi}{4} ) and its coterminal angles.", "---", "### Key Trigonometric Values: ( \sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} )", "More precisely, since sine is an odd function, we know:", "[\n\sin\left(-\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}\n]", "Thus, one solution to ( \sin(c) = -\frac{1}{\sqrt{2}} ) is:", "[\nc = -\frac{\pi}{4} + 2k\pi \quad \ ext{(for any integer } k\ ext{)}\n]", "But sine is periodic with period ( 2\pi ), so the general solution includes all angles coterminal with ( -\frac{\pi}{4} ), plus inversion due to sine’s symmetry.", "---", "### General Solutions: All Solutions to ( \sin(c) = -\frac{1}{\sqrt{2}} )", "The sine function equals ( -\frac{1}{\sqrt{2}} ) at angles where the reference angle is ( \frac{\pi}{4} ), and sine is negative in the third and fourth quadrants. Therefore, the general solution is:", "[\nc = \frac{5\pi}{4} + 2k\pi \quad \ ext{or} \quad c = \frac{7\pi}{4} + 2k\pi \quad (k \in \mathbb{Z})\n]", "Alternatively, expressed using a single reference angle:", "[\nc = \pi + \frac{\pi}{4} + 2k\pi = \frac{5\pi}{4} + 2k\pi\n]\n[\nc = 2\pi - \frac{\pi}{4} + 2k\pi = \frac{7\pi}{4} + 2k\pi\n]", "---", "### Understanding the Graph and Behavior", "Graphically, ( \sin(c) = -\frac{1}{\sqrt{2}} ) corresponds to horizontal lines crossing the unit circle at points where the y-coordinate is (-\frac{1}{\sqrt{2}}). These occur repeatedly every ( 2\pi ) radians, confirming the periodic nature of sine.", "---", "### Applications and Relevant Concepts", "Understanding this transformation is useful in:", "- Solving trigonometric equations in calculus and physics\n- Analyzing waveforms and harmonic motion\n- Computing angles in coordinate geometry and complex numbers", "Knowing that ( \sqrt{2} \sin(c) = -1 ) transforms directly to ( \sin(c) = -\frac{1}{\sqrt{2}} ) allows precise set-up and faster solution derivation.", "---", "### Conclusion", "The equation ( \sqrt{2} \sin(c) = -1 ) simplifies elegantly to ( \sin(c) = -\frac{1}{\sqrt{2}} ), linking directly to key trigonometric identities and standard angles. Recognizing this equivalence supports accurate solving, deeper conceptual understanding, and effective application in mathematical and physical modeling.", "---", "Keywords for SEO:\nsolve \( \sqrt{2} \sin(c) = -1 \),\nsine equation solutions,\n\( \sin(c) = -\frac{1}{\sqrt{2}} \),\ntrigonometric identities,\nunit circle sine values,\nsine graph periodicity,\ntrigonometry problem solving", "---"]