Download Chapter 13анаTrigonometric Ratios and Functions Pythagorean

Chapter 13 ­ Trigonometric Ratios and Functions Pythagorean Theorem: Use it to find the missing side, in order to solve for missing angles, if it doesn't work out as whole number, leave as a square root Inside Angle measures: Interior angles of triangles add to be 180, can be used to find missing angles SOH/CAH/TOA Trig Functions Reciprocal Trig Functions (Sally can tell, Oscar has a hat on always) TO SOLVE: * Draw a picture and label all given sides and angles * Determine what angle you are trying to find and what info is given * Use the appropriate reciprocal to solve * Set up as a proportion using given info and x for the missing info * Theta ­ unknown angle, will be marked in a diagram * Can't have a square root sign in denominator (multiply top and bottom by square root sign, value stays in denominator), that's when you'd use reciprocal functions Angle of depression: Looking down from the horizon, the theta angle is formed by the outside of the triangle and the horizon Angle of elevation: Looking up from the ground, or eye level Special Triangle Ratios 45 ­ 45 ­ 90 Divide by square root of 2 to find short sides 30 ­ 60 ­ 90 Mutliply short side by square root of 3 to find long side (Set up in proportions, cross multiply, solve for missing letter. ) Degrees and Radians An angle is in standard position if its: ● initial side is on the positive x axis ● vertex at the origin ● measured counterclockwise to its terminal side ( ending position) * Coterminal angles share the same terminal side (end in same point) * To find terminal angles, add 360 and subtract 360. * Counterclockwise is a positive angle, clockwise is a negative angle. * Infinite number of angles (­) and (+) if you're given angle is theta. * May have to keep subtracting to get to a (­) value if the number is high enough. Radians: ● 1 radian is the angle required to map out one radius of arc length on a circle ● The unit of measure that tells how far around the circle we've traveled at a certain degree compared to the radius ● Radians are written as fractions to describe the fraction of the circle you've traveled as a unit in terms of pii ● Cirumference = 2pii * r ● Radius = 1 on the unit circle ● 180 degrees = pii ● 360 circle = 2pii ● Can find coterminal angles using radians ­ subtract or add 2pii/1 instead of 360, will need to find common denominators in order to add or subtract the fractions. Keep going to get (­) or (+) values. Changing from Radians to Degrees, and Degrees to Radians: *Set up as proportions, multiply straight across, reduce fractions. *Can replace pii, with 180 degrees, to complete steps. Examples: UNIT CIRCLE DIAGRAM Radius = 1 Cosine = x/r
Sec = r/x Sin = y/r
csc = r/y Tan = y/x
cot = x/y When writing trig ratios, determine which degree to use on unit circle, and use correct values to build proportions. Simplify proportions, remembering you can’t have a squared root in the denominator. Starting in 1st quadrant counterclockwise, ASTC, All scholars take calculus. Tells which value is positive in each quadrant. BIG IDEA: X values are cosine, Y values are sin (cos, sin). Reference Angles: The acute angle formed by the terminal side of theta and the x ­ axis. Draw the given angle, determine which quadrant it ends in. (+) counterclockwise, (­) clockwise If the problem is in radians, keep in radians. Quadrant
How to find Reference Angle 1
Same as given 2 180 ­ given 3 Given ­ 180 4 360­given 2
1 3
4 14 ­ 1: Periodic Intervals Curves are really just stretched out unit circles. Remember y is sin, x is cos. Visualize x and y values traveling around circle. Amplitude = 1 ● Half the range Absolute value of a (like a distance) ­ amount of sway between x axis. ● ( will be the value in front of cos/sin...if no number given, use 1) ● Creates boundary lines above and below x axis Period = 2pii/b ● Distance required to repeat the wave ­ distance between each wave point ● 2 pii/ b ­ repeating patterns on x axis ● (will be the value after sin/cos...if no number given, just use 2pii) ● y ­ Sin theta ( 0, y) Starts at 0, start down if (­), up if (+) ● x ­ Cos theta (x, 0) Starts at amplitude, start down if (­), up if (+) y = a sin 60 theta
y = a cos b theta Ex. Graph y = 2 sin 3 theta Graphing a curve: 1. Start with curve 2. Figure out amplitude (2) ­ mark on curve 3. Figure out period (2pii/ 3) Shrinks the value, because you are dividing by a number 4. Put unit circle markings on the curve in order by dividing period into 1/2, 1/4 etc Can determine amplitude and period from curves: 1. Look for distance up and down y axis: amplitude 2. Determine period ­ take highest number on curve, divide it by b value given ( number after cos/or sin). * Periods for cos x or sin x are always 2pii. Ex. y = 2 cos 1/4pii x ­ change 1/4 pii to pii/4 Amp: 2 Period is 2pii/pii 4 ­ multiply by reciprocal, solve. Period is 8. Graph curve. 14.2 Shifting and reflecting ( not on exam) y = a sin [ b ( x ­ h ) ] + k ● h value moves it right or left ­ opposite of what you think ● k value moves it up or down ­ move exactly the way you think ● if a negative or <0 , it is a reflection over x axis Ex. Shifting left or right (opposite of what you think) Graph y = 2 sin(x + pii/4) Amp = 2 Period = x , which is 2pii Shift left pii/4 To make curve: 1. Move left 4 2. Go up, down, etc 3. Mark amplitude 4. Mark period 5. Make markings on curve Ex. Shifting up or down (what you think) y = 1/2 cos (2x) + 3 Amp = 1/2 Period = 2 pii/2 or pii Shift = up 3 units , changes the baseline ­ no longer along 0 Ex. Reflecting y = ­2 cos ( x + pii/6) Amp = 2 Curve = reflects over x axis Period = 2pii Shift = left, pii/6 Graph = shift left ­pii/6, end at 2pii Fill in other numbers = take highest number and subtract (period plus shift) to find other points along x axis Ex. y = ­ 1/2 sin 2 ( x ­ pii/4) + 1 amp = 1/2 Period = 2pii/2 or pii Shift = right pii/4 Shift = up 1 ­ x axis is going to be on 1, not zero Fill in other numbers = Ends at 2pii, subtract the (period plus shift) to find other points. BIG IDEA: y = cos x = sin ( x + pii/2) y = cos x = cos ( x ­ pii/2) 13.5 ­ 13.6 Law of Sines/ Law of Cosines ● Law of cosines when given SAS, SSS ● Law of sines when given ASA AAS ● Use this for triangles that are not 90 degree right triangles ­ cannot do SOHCAHTOA Law of Sines: Used to find missing sides Sin A = Sin B = Sin C a b c Steps to solve: 1. Label all values and record the given information (make a chart that includes all 6 parts) 2. Set up proportion of law of sines with given info to find missing info 3. Cross multiply and divide to solve Example: Solve triangle PQR in which ∠ P = 63.5° and ∠ Q = 51.2° and r = 6.3 cm. Solution: First, calculate the third angle. ∠ R = 180° – 63.5° – 51.2° = 65.3° Next, calculate the sides. ∠ R = 65.3°, p = 6.21 cm and q = 5.40 cm Law of Cosines: Used to find missing angles a 2 = b 2 + c 2 – 2bc cos A b 2 = a 2 + c 2 – 2ac cos B c 2 = a 2 + b 2 – 2ab cos C Steps to solve: 1. Label all values and record the given info ( make a chart including all 6 parts) 2. Use the equation for the Angle you are given 3. Plug in number values to solve ­ punch into calculator, take square root 4. Once you have found first side, you can use law of sines to find rest of sides. *Avoid finding largest angle (opposite largest side), as a calculator can't put out an obtuse angle. 5. Will need to use Sin ^­1 (inverse sin) to find other side. 6. To find third side, add the two and subtract from 180. Example: Solve triangle PQR in which p = 6.5 cm, q = 7.4 cm and ∠R = 58°. Solution: Using the Cosine rule, r2 = p2 + q2– 2pq cos R r2 = (6.5)2 + (7.4)2– 2(6.5)(7.4) cos58° = 46.03 r = 6.78 cm Using the Sine rule, ∠Q = 180° – 58° – 54.39° = 67.61° ∠P = 54.39°, ∠Q = 67.61° and r = 6.78 cm When given SSS use: −1
Cos (a 2 ­b 2 ­c 2 ) = A ­2bc −1
Cos (b 2 ­a 2 ­c 2 ) = B ­2ac −1
Cos (c 2 ­ b 2 ­ a 2 ) = C ­2ab 1. Choose what you are trying to solve, choose formula for correct angle (find largest 1st) 2. Plug in given information for the correct formula 3. Solve 4. Once you have one angle, then you can do Law of sines to find all missing info Example: In triangle ABC, a = 9 cm, b = 10 cm and c = 13 cm. Find the size of the largest angle. Solution: The largest angle is the one facing the longest side, i.e. C. c2 = a2 + b2– 2ab cos C = 0.067 ∠C = 86.2°