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Published byDamian Perkins Modified over 8 years ago
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Math Review Egr 102-2101
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Trigonometry Consider a right triangle as shown with sides a and b and hypotenuse c. B c a A b
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Angle A is complementary to angle B A + B = 90⁰ a 2 + b 2 = c 2 --- Pythagorean Theorem Sin A = a/c (opposite side/hypotenuse) Cos A = b/c (adjacent side/hypotenuse) Tan A = a/b (opposite side/adjacent side) Sin 2 A + cos 2 A = 1 [Prove this!]
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Sin 2 A + cos 2 A = (a/c) 2 + (b/c) 2 = (a 2 + b 2 )/c 2 = 1 By the Pythagorean Theorem
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Sin A = cos B Sin A = a/c Cos B = a/c So Sin A = cos B if A and B are complementary angles.
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Sin 0 = 0 Cos 0 = 1 Sin 90 = 1 Cos 90 = 0 Sin 180 = 0 Cos 180 = -1 Sin 270 = -1 Cos 270 = 0 Sin 360 = 0 Cos 360 = 1 Sin and Cos vary in a sinusoidal manner but are out of phase with one another by 90 degrees.
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c a A b If the value of c = 2.0 and angle A is 30⁰, Then a = c sin 30 = 2.0 sin 30 = 2.0 x 0.5 = 1.0 Then b = c cos 30 = 2.0 x 0.866 = 1.732 And this checks out by the Pythagorean Theorem
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Exponentials a 1 = a a 2 = a x a a 3 = a x a x a a 1 x a 2 = a (1+2) = a 3 a 1 / a 2 = a (1-2) = a -1 = 1/a 5 3 x 5 4 = ? 5 3 / 5 4 = ?
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5 3 x 5 4 = 5 7 5 3 / 5 4 = 5 -1
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Logarithms There are basically two types of logarithms we use: Common logarithms or logs to the base 10 Natural logarithms or logs to the base e Common logs are called log Natural logs are called ln
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Working with Logs Log ab = log a + log b Log a/b = log a – log b Log a n = n log a Log a (1/n) = (1/n) log a
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log (2)(3) = log 6 = 0.7782 log 2 = 0.3010 log 3 = 0.4771 log 2 + log 3 = 0.7781 log (2/3) = log (0.667) = -0.1759 log 2 – log 3 = 0.3010 – 0.4771 = -0.1761 rounding off errors
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log 2 3 = 3 log 2 = 3(0.3010) = 0.9030 log 8 = 0.9031 log 8 (1/3) = log 2 (1/3)log 8 = (1/3)(0.9031) = 0.3010
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Exponentials e -ax = 1/e ax ln (e -ax ) = -ax
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