# Common Trig Ratios :: Degrees

## Finding the exact values of sine, cosine and tangent of special angles.

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Solve these trigonometric equations to find the solution in the given domain. Non exact answers should be given to three significant figures.

 $$\tan(x) = 1\\ 90° \lt x \lt 270°$$ $$x=$$ ° $$\sin(x) = 0.5\\ 90° \lt x \lt 180°$$ $$x=$$ ° $$x = \cos^{-1}(0.5)\\ -90° \lt x \lt 0°$$ $$x=$$ ° $$x = \tan^{-1}(0)\\ 90° \lt x \lt 270°$$ $$x=$$ ° $$\sin(x) = \frac{ \sqrt{3}}{2} \\ 90° \lt x \lt 180°$$ $$x=$$ ° $$\cos(x) = \frac{ \sqrt{3}}{2} \\ 0° \lt x \lt 270°$$ $$x=$$ ° $$\frac{6}{\cos(x)} = 10\\ 180° \lt x \lt 360°$$ $$x=$$ ° $$2(3\sin(x) - 1) = \frac{1}{4}\\ 450° \lt x \lt 540°$$ $$x=$$ ° $$7\tan(x) + 4 = 5\tan(x) \\ 180° \lt x \lt 360°$$ $$x=$$ ° $$5\tan(x) - 3 = 1 + 2\tan(x) \\ -90° \lt x \lt 90°$$ $$x=$$ ° $$2\sin(x) + 1 = 2.1\\ 360° \lt x \lt 450°$$ $$x=$$ ° $$5\cos(x) - 2 = \frac{1}{3} \\ -180° \lt x \lt 0°$$ $$x=$$ ° $$\frac{5}{\sin(2x)} = 8\\ -45° \lt x \lt 45°$$ $$x=$$ ° $$\tan^2(3x) = 5\\ 100° \lt x \lt 150°$$ $$x=$$ ° $$6\cos^2(x) = 1 - \cos(x) \\ -180° \lt x \lt -90°$$ $$x=$$ °
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## Description of Levels

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Surds - Make sure you understand what surds are before starting the levels below.

Level 1 - Find exact trig values for special angles up to and including ninety degrees

Level 2 - Find the indicated lengths by solving trigonometric questions with exact solutions

Level 3 - Mixed questions on exact trig values of special angles up to and including ninety degrees

Level 4 - Find exact trig values for angles between ninety and three hundred and sixty degrees

Level 5 - Solving trigonometric equations with given domains

Radians - A similar set of exercises worked in radians.

If Then - Find one trig ratio if given another from the same triangle.

Exam Style Questions - A collection of problems in the style of GCSE or IB/A-level exam paper questions (worked solutions are available for Transum subscribers).

More Trigonometry including visual aids, investigations and self-marking exercises.

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