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• 1. MATH: TRIGONOMETRY Section V DePaul Math Placement Test
• 2.
• Basic Trigonometric Functions
Most of the trigonometry on the math placement test addresses the different parts of a right triangle and the relationships among these different parts. The three basic trigonometric functions— sine, cosine, and tangent —are the tools that define these relationships. Given the measure of one of the non-right angles in a right triangle, you can use these tools of trigonometry to find the characteristics of the triangle. If you are given the measure of one of the non-right angles and one of the sides, you can find all the values of the right triangle.   Basic Functions and the Right Triangle If you know the measure of one of the non-right angles in a right triangle, the trigonometric functions tell you the ratio of the lengths of any two sides of the triangle. In the right triangle below, one acute angle is labeled and the sides of the triangle are labeled hypotenuse, opposite, and adjacent, according to their position relative to the angle of measure .
• 3.
• Basic Functions and the Right Triangle
If you know the measure of one of the non-right angles in a right triangle, the trigonometric functions tell you the ratio of the lengths of any two sides of the triangle. In the right triangle below, one acute angle is labeled and the sides of the triangle are labeled hypotenuse, opposite, and adjacent, according to their position relative to the angle of measure Sine: The sin of an angle is the ratio of the side opposite the angle to the hypotenuse. sin =opposite/hypotenuse Cosine: The cos of an angle is the ratio of the side adjacent the angle to the hypotenuse. cos= adjacent/hypotenuse  Tangent: The tan of an angle is the ratio of the side opposite the angle to the side adjacent to the angle. tan= opposite/adjacent
• 4.
•   Reciprocal functions
• There 3 more function other than sin, cos and tan which are related to these three functions:
• The cosecant: csc( A ) is the reciprocal of sin( A ) , i.e. the ratio of the length of the hypotenuse to the length of the opposite side:
• csc=(1/sin) = hypotenuse/opposite
• The secant: sec( A ) is the reciprocal of cos( A ), i.e. the ratio of the length of the hypotenuse to the length of the adjacent side:
• The cotangent: cot( A ) is the reciprocal of tan( A ) , i.e. the ratio of the length of the adjacent side to the length of the opposite side:
• 5.
• Basic Trigonometric Functions
Angles Larger than 90 0 and the Basic Functions Angles in a right triangle can never be larger than 90º, since the sum of all three angles must equal 180 0 . But in the math placement test, you may occasionally run into angles that are larger than 90 0 . It is often more intuitive to think of these in terms of the coordinate plane rather than in terms of a triangle.   Angles and quadrants Below are pictured four angles in the coordinate plane. The first is the acute angle we’ve already covered in this chapter; the next three are all larger than 90 0 . The four quadrants of the coordinate plane become very important when dealing with angles that are larger than 90 0 . Each angle larger than 90 0 can be “simplified” by looking at it in the context of its own quadrant.
• 6.
• By reconsidering each angle based on its relationship to the x-axis as shown on the left hand size, and taking 30 0 as an example, becomes clearer it becomes clear that each of the original angles can be treated as a reoriented 30 0 angle.
• In other words, a 210 0 angle is just the same as a 30 0 angle except that the 210 0 angle is in the third quadrant.
• In terms of the basic trigonometric functions, this means that the value of a 210 0 angle is the same as the value of a 30 0 angle, except that the sign of the trigonometric function differs based on the quadrant that the angle is in.
• Depending on the quadrant of the coordinate plane in which an angle resides, the values of the trigonometric properties of that angle will be either positive or negative.
•
• 7.
You should know the proper sign for each quadrant in an indirect way, meaning that it’s unlikely that you’ll have to do any heavy calculating when dealing with this topic. Instead, you might find a question such as: Q. If the value of sin – is .5 , what is the value of sin ? A. The first thing you should notice is that - and have the same magnitude, even if they have different signs. This means that the magnitude of sine for – and will be the same. Immediately you should understand that sin must equal either .5 or –.5. To figure out which of these values is right, you have to decide what quadrant angle resides in. Based on the graph of the sine function or from the above chart, you can see that the sine function has a positive value in quadrants I and II, and negative values in quadrants III and IV. Since sin– is equal to a positive number, .5, you know that – must represent an angle in quadrant I or II. Since angle is simply the reflection of – across the x -axis, you can see that angle must be in either quadrant III or IV. The value of sin must be negative: – . 5 is the right answer.
• 8.
• Solving Right Triangles
• One of the most important applications of trigonometric functions is to “solve” a right triangle. By now, you should know that every right triangle has five unknowns:
• the lengths of its three sides and the measures of its two acute angles.
• Solving the triangle means finding the values of these unknowns.
• You can use trigonometric functions to solve a right triangle if you are given either of the following sets of information:
• The length of one side and the measure of one acute angle
• The lengths of two sides
•
• 9.
• Solving Right Triangles
•   Usually the common way of question for this section is that
• Q. Given: One Angle and One Side
• The right triangle below has an acute angle of 35º and a side of length 7.
• To find the measure of the other acute angle, just subtract the measures of the other two angles from 180º:
• A.
• L B= 180-90-35=55 o
•
• To find the lengths of the other two sides, use trigonometric
• functions relating the given angle measure to the given side length.
• In this question, you are given the measure of one angle and the length of the side opposite that angle, and two trigonometric functions relate these quantities. Since you know the length of the opposite side, the Sin will allow you to solve for the length of the hypotenuse. And tan will let you solve for the length of the adjacent side . sin 35 o = 7/c; tan 35 o = 7/b.
• c= 7/sin 35 o =12.2 ; b = 7/tan(35 o ) = 10.0. {To find the values of trigonometric functions use you calculator. For further instruction see section below the slides.}
• 10.
• Solving Right Triangles
Here is another Example: Q. Given: Two Sides The right triangle below has a leg of length 5 and a hypotenuse of length 8. First, use the Pythagorean theorem to find the length of the third side: 2   b= = 6.2   Next, use trigonometric functions to solve for the acute angles:   sin A = 5/8 cos B = 5/8 Now you know that sin A = 5 ⁄ 8 , but you are trying to find out the value of L A, not sin A . To do this, you need to use some standard algebra and isolate L A. In other words, you have to find the inverse sine of both sides of the equation sin A = 5 ⁄ 8 . Carrying out this operation will tell you exactly which angle between 0 o and 90 o has a sine of 5 ⁄ 8 . sin -1 (sin A) = sin -1 (5/8) L A=38.7 o   You can solve for L B by using the cos –1 button and following the same steps.(51.3 o ).
• 11.
• General Rules of Solving Right Triangles
• You have seen two of the different paths you can take to solve a right triangle. The solution will depend on the specific problem, but the same three tools are always used:
• The trigonometric functions
• The Pythagorean theorem
• The knowledge that the sum of the angles of a triangle is 180 o
• Remember: There is no “right” way to solve a right triangle. One way that is usually wrong, however, is solving for an angle or a side in the first step, approximating that measurement, and then using that approximation to finish solving the triangle. This approximation will lead to inaccurate answers, which in some cases might mean that your answer will not match the answer choices.
• 12.
• Trigonometric Identities
A trigonometric identity is an equation involving trigonometric functions that holds true for all angles. These identities are commonly called Pythagorean identities, because they come from the Pythagorean theorem. tan = Sin / Cos sin 2 + cos 2 = 1 You will have to by asking you to simplify a complex expression. Answering these questions has more to do with memorizing the identities and being good with algebraic substitution than it does with the theoretical concepts of trigonometry. For example: What is (cos tan ) / (sin – cos 2 )? To solve a problem like this, use the trigonometric identities to simplify the trigonometric into sines and cosines. After you have simplified the expression using the identities, it is quite likely that the expressions will simplify further due to the canceling of terms. The simplification of the expression in the example question proceeds as follows: (Cos X tan )/Sin – Cos 2 = [(Cos X Sin ) /Sin X Cos ] - Cos 2 = 1-Cos 2 = Sin 2   Simplifying the mess given to you by the problem, you get sin 2 .
• 13.
• Unit Circle
The unit circle is a circle whose center is the origin and whose radius is 1. It is defined by equation x 2 + y 2 = 1. The most useful and interesting property of the unit circle is that the coordinates of a given point on the circle can be found using only the measure of the angle. Any radius of the unit circle is the hypotenuse of a right triangle that has a (horizontal) leg of length cos and a (vertical) leg of length sin . The angle is defined as the radius measured in standard position. These relationships are easy to see using the trigonometric functions:   sin = y/1 =y cos = x/1 =x tan = y/x As you can see, because the radius of the unit circle is 1, the trigonometric functions sine and cosine are simplified: sin = y and cos =x . This means that another way to write the coordinates of a point (x, y) on the unit circle is (cos , sin ), where he measure of the angle in standard position whose terminal side contains the point.
• 14.
• Unit Circle
The Unit Circle and Important Angles Using the unit circle makes it easy to find the values of trigonometric functions at quadrantal angles. For example , a 90 o rotation from the positive x -axis puts you on the positive y -axis, which intersects the unit circle at the point (0, 1). From this, you know that (cos 90 o , sin 90 o ) = (0, 1). Here is a graph of the values of all three trigonometric functions at each quadrantal angle: There are a few other common angles besides the quadrantal angles whose trigonometric Function values you should already know. Listed on the bottom-are the values of sine, cosine, and tangent taken at 30 o , 45 o and 60 o .You might recognize some of these values from the section on special triangles.
• 15.
• Radians are another way to measure angles. Sometimes radians will be used in questions, and other times you may choose to use them since they are sometimes more convenient than degrees.
• A degree is equal to 1 / 360 of a circle, while a radian is equal to the angle that intercepts an arc of the same length as the radius of the circle. In the figure below, arc AB has length r , and the central angle measures one radian.
• When converting between the two measurement systems, use the proportion:
• 1 degree/360 = 1 radian/2 π
• which can be simplified to:
• 1 degree/180 = 1 radian/ π
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•
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• 16.