WEBVTT

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Here is the problem: Find arcsin(negative the square root of 2 over 2 and every angle in [0°, 360°) with sin x equals negative the square root of 2 over 2.

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Find the angle whose sin is negative the square root of 2 over 2: the principal value of arcsin, every solution of sin(x) equals negative the square root of 2 over 2 in one turn, and the general solution.

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Principal value of arcsin, taken in the range [−pi/2, pi/2]. negative 45° is a special angle, so the answer is exact.

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Sine is also positive or negative in a second quadrant, so use x and 180° minus x.

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Solutions in one full turn, 0° less than or equal to x less than 360°.

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General solution, n any integer.

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So the answer is arcsin(negative the square root of 2 over 2 equals negative 45° (-pi/4); solutions in [0°, 360°): 225°, 315°; general: x equals negative 45° plus 360°n or x equals 225° plus 360°n.

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An inverse trigonometric function returns one angle, the principal value. Because the trigonometric functions repeat, the equation sin(x) equals negative the square root of 2 over 2 has infinitely many solutions; the calculator lists those in one turn and the general pattern.

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Substituting each solution back: sin(225°) approximately negative 0.7071067812; sin(315°) approximately negative 0.7071067812.