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On this page we hope to clear up problems you might have with trig. functions in equations and triangles (law of sines, etc.).  Scroll down or click any of the links below to gain a better understanding of trigonometry in equations and triangles. Inverses of trig. ratios Trigonometric equations Triangles Law of sines Law of cosines Quiz on Equations and Triangles The inverses of the trigonometric ratios are denoted in the three different ways, which are shown below. x = sin y y = sin-1 x y = arcsin x Arcsin x is a number whose sine is x. Some rotations do not have values for the inverse trig. functions.  100o is an example that does not have an arcsin.  The ranges of the inverse functions are listed below. y = Arcsin x [-((PI)/2), (PI)/2] y = Arccos x [0, (PI)] y = Arctan x [-((PI)/2), (PI)/2] Example: 1. Problem: Find Arcsin ((SQRT(2))/2). Solution: Using a unit circle like the one pictured below, you can see that there is only one angle in the Arcsin's range that has a sine of (SQRT(2))/2. That angle is (PI)/4. Arcsin (SQRT(2))/2 = (PI)/4 Back to Top  Trigonometric Equations are equations that contain expressions such as sin x.  You solve a trigonometric equation in the same way as any other equation.  Example: 1. Problem: Solve for x: 2sin x = 1. Solution: Solve for sin x by dividing by 2. sin x = (1/2) The solutions are any angles that have a sine of (1/2).  (This could also be written as arcsin (1/2).) Using a unit circle, you can locate the angles where the sine is (1/2). There are two angles on the unit circle, ((PI)/6) and (5(PI))/6. Also, if you add any multiple of 2(PI) to the angles you also have valid solutions (because ((PI)/6) + 2(PI) also has a sine of (1/2)). ((PI)/6) + 2k(PI) and (5(PI)/6) + 2k(PI) where k is any integer. Back to Top  Triangles are useful when you need to find unknown lengths and heights.  Example: 1. Problem: Find length b. Solution: We see that we know the length of the hypotenuse and the measure of angle A. We also see that we need to find the length of b. b is adjacent to angle A. By definition, cosine = adjacent over hypotenuse. Therefore, we write an equation using the cosine. cos 19o = b/70 Multiply each side by 70 to isolate the variable b. 70(cos 19o) = b Use a calculator to find the cos 19o. b = 66.19 Back to Top  Before now, we have been able to use the trigonometric functions to solve right triangles.  This doesn't do much good because there are many times triangles will not be right triangles.  The Law of Sines is one way that we can solve oblique triangles.  It is listed below. a b c ----- = ----- = ----- sin A sin B sin C When you know any two angles and any side of a triangle, you can use the law of sines to solve the triangle.  Example: 1. Problem: In triangle ABC, a = 4.56, A = 43o, and C = 57o. Solve the triangle. Solution: First, sketch the triangle and include any given information. Here's our idea of a sketch: Find angle B. Angle B = 180o - (43o + 57o) = 80o Now, we use the law of sines to find the other sides lengths. c a ----- = ----- sin C sin A a(sin C) c = -------- sin A Plug in any known information. 4.56(sin 570) c = ------------- sin 43o Use a calculator to find the sines. 4.56(.8387) c = ----------- .6820 c = 5.61 Now solve for b. b a ----- = ----- sin B sin A a(sin B) b = -------- sin A Plug in any known information. 4.56(sin 80o) b = ------------- .6820 Use a calculator to find the sines. 4.56(.9848) b = ----------- .6820 b = 6.58 There is another case when you can use the law of sines.  When you know two sides and an angle opposite one of the sides, the law of sines can be used.  However, with this case, you have to be aware that there might not be a solution, or there may be two!  One solution is also possible.  Example: 2. Problem: In triangle ABC, a = 15, b = 25, and angle A = 47o. Solve the triangle. Solution: Start out by looking for the measure of angle B. a b ----- = ----- sin A sin B b(sin A) sin B = -------- a Plug in any known information. 25(sin 47o) sin B = ----------- 15 Use a calculator to find the sine. 25(.7314) sin B = --------- 15 sin B = 1.219 Since an angle cannot have a sine greater than 1, there is no solution for this triangle. Back to Top  You can also solve triangles that are not right triangles using the Law of Cosines.  There are three rules that make up the law of cosines, but you only need to memorize one because the other two can be obtained by changing the letters (put b in place of a, for example).  All three rules are listed below. a2 = b2 + c2 - 2bc(cos A) b2 = a2 + c2 - 2ac(cos B) c2 = a2 + b2 - 2ab(cos C) The law of cosines can be used in two different cases.  First, if you know the lengths of all three sides of a triangle, you can use the law of cosines.  Secondly, you can use the law of cosines when two sides and the included angle are known.  Example: 1. Problem: In triangle ABC, a = 24, c = 32, and angle B = 115o. Solve the triangle. Solution: We know two sides. Find the third using the law of cosines. b2 = a2 + c2 - 2ac(cos B) Plug in any known information. b2 = 242 + 322 - 2 * 24 * 32(-.4226) b2 = 2249 Solve for b by taking the square root of each side. b = SQRT(2249) = 47.4 Using the law of sines, angle A = 27.32o. Since we know that triangles are made of three angles that sum to 180o, we can find the measure of angle C by setting up an equation. Angle C = 180o - 115o - 27.32o = 37.68o Back to Top  Take the Quiz on equations and triangles.  (Very useful to review or to see if you've really got this topic down.)  Do it!