DC Trigonometry Final Exam Concepts

example of "use a geometric series to find the rational number represented by the repeating decimals"

3.17 (.17 repeating) -> 3 + .17 -> r = .01 -> .17/(1-.01) -> .17/.99 -> 17/99 -> 3/1 * 99/99 -> 297/99+17/99 = 314/99

find the exact values of the 6 trig function of the angle theta given the angle from standard position. given: (5,3)

begin by drawing a graph and plotting the point. create a right triangle and label the sides. use pythagorean theorum: 3^2+5^2=c^2. c=sqrt(34) which is the hypotenuse or the missing side. the 6 trig functions are SOH CAH TOA and CSC SEC COT. list all possible values. remember to rationalize fractions with a square root in the denominator. if you are given a problem where theta bisects a quadrant it will be a 45º 45º 90º triangle.

express the sum in terms of summation notation and find the sum

for this one we are given a sequence of numbers that are being added in a particular pattern. this is either arithmetic or geometric. arithmetic is when something is being added over and over and geometric is when something is being multiplied over and over. the number being added or multiplied is the common difference or "d". the formula for the arithmetic sum is S = n/2*(a_1+a_n) in this formula you will need to find "n" or the amount of terms there are. the formula for an arithmetic series, which we can use to find n is: a_n=a_1+(n-1)d where a_n is the last term. solve for n and plug n in for the arithmetic sum equation.

use a geometric series to find the rational number represented by the repeating decimals

split the repeating number from its rational part, and write them as a fraction (rational over one). use the formula: a_1/(1-r) to write the repeating part as a fraction. a_1 = repeating number and r = the number of decimal places in your repeating number ending with a one. after calculating the formula, move the decimal places over as many times as needed on the numerator and denominator in order to get two whole numbers. after multiplying the denominator to the top and bottom of the rational number, add them. to check your answer, use a scientific calculator and the original number should appear.

find the period, phase shift, y-intercept, and three x-intercepts. then graph an interval of at least 2pi. be sure to label all x-intercepts and asymptotes. given: f(x)=tan(2x-pi) and a graph

steps: 1. find period (pi/b) 2. find intervals (period/2) & label x-axis 3. find phase shift 4. plot x-intercept & alternate intercepts & asymptotes y=a tan b(x-h)+k b is what is in front of the x-term. amplitude or a is in front of the identity, but tangent doesn't have any. h is horizontal shift and k is the vertical shift. since the b value was distributed, we need to divide both terms by 2 so instead of -pi we now have -pi/2. 1. pi/2 2. pi/4 (check x-axis, the points should be at every pi/4 which may be in between values) 3. -h is the phase shift. pi/2 (take the opposite +-) 4. phase shift is the first x-int, and the rest alternate between asymptotes and intercepts. graph from left to right, starting on the bottom when positive. if negative, start on the top, going left to right. do not touch the asymptotes. the y-int is the point where it crosses the y-axis.

suppose a is a unit vector. find the exact value of x that satisfies the condition

we are given "a={x,2/3} in quadrant 1". a unit vector means it's length or magnitude is 1. we go to the right x units and up 2/3 units, and this creates a right triangle (it helps to draw this one too). for this we can use pythagorean theorum. x^2+2/3^2=1^2 then we get x^2=5/9 which we need to take the square root of. this gives us x=+-sqrt(5)/3 do not forget the +-!!!! now we can look back at the graph to determine if we should answer with the positive or the negative. since the graph is in quadrant 1, it's positive.

find the exact value of the dot product of the two vectors

we are given a={2,3} and b={-5,-7}. we need to multiply the two vectors together. 2(-5) and 3(-7) which is -10 and -21 which need to be added together to get -31.

find the value of x that makes the vectors orthogonal.

given: {2,-7} and {x,4}. to find something is orthogonal, the dot product should equal 0. 2(x)+(-7)(4)=0. solve for x and x=14.

find the component of vector a onto vector b (looks like: comp_(b)a.) given: a={2,-4} and b={6,8}

take the dot product of a and b divided by the magnitude of b. a*b/magb. 2(6)+-4(8)=-20 and magb= sqrt(6^2+8^2)=10. -20/10=-2.

suppose angles A and B are angles lying in the 2nd quadrant. find the indicated value under the given circumstances. given: cosA=-4/7 and cscB=2 *sin(A+B)=

begin by graphing the fraction, and forming a right triangle. be sure to use SOH CAH TOA when labeling the sides. draw a second graph for B. since cscB=2, sinB=1/2 because it's reciprocal. sin(A+B) is the same as sinAcosB+cosAsinB. do Pythagorean theorem on each triangle to find the missing sides. then we find the fractions we need to plug into the problem. (sqrt33/7)(-sqrt3/2)+(-4/7)(1/2)=((-3sqrt11)-4)/14.

find the exact value of the area of triangle ABC

use the formula A=sqrt[s(s-a)(s-b)(s-c)] where s is the semi perimeter. to find the semi perimeter, add up all three sides and divide by 2. then plug all of the values into the equation to get the area.

approximate the value given the question below to the nearest hundredth. given: a person looks up to see a bird in a tree 30 feet in front of them. if the person's eyes are at a height of 5 feet and the angle of elevation to the bird is 55º, find the height of the bird above ground level.

draw a picture showing the given values in the triangle, and the values in the rectangle below the triangle. this is a right triangle on top, and the base of the rectangle is equal to the base of the triangle. we can use SOH CAH TOA to find the missing side. in this case we will use tan55=x/30. multiply 30 on both sides to get x by itself. x=30tan55º which is approximately 42.84. we only have a part of the height of the tree, so now we need to add the lower half which we know is 5 feet. 42.84+5= 47.84 feet.

verify trigonometric identities given: secx/(1-secx)-secx/(1+secx)=-2csc^2x

first of all, we need to pick one side and simplify it down to the other side. it will be easier if we simplify the bigger, messier side. we can combine the two fractions on the messy side by getting a common denominator. multiply the top and bottom of EACH fraction with the denominator of the opposite fraction. combine them into a single fraction and simplify by distributing. if anything in the numerator or denominator cancels out, be sure to double check. recall pythagorean identities. the denominator in this problem, at this point is (1-sec^2x) which is equal to -tan^2x. we now have (2sec^2x)/(-tan^2x). these can be simplified further using the identities that form them. (ex. tan is sin/cos) and sec is 1/cos) leave the ^2x in the identities. instead of dividing by the fraction, flip the fraction and cross multiply it to the numerator. sometimes identities will cancel. the final answer for this problem is -2csc^2x which is equal to the other side of the problem, so it's correct.

find the exact value of cos theta between vectors a and b

given: a={4,1} and b={3,-3}. formula is cos theta=(dot product or a*b)/(magnitude a*magnitude b). 4(3)+1(-3)=9. use pythagorean theorum for the magnitudes. sqrt(4^2+1^2)=sqrt(17). sqrt(3^2+(-3)^2)=3sqrt(2). 9/sqrt(17)(3sqrt(2))=9/(3sqrt(34))=3/sqrt(34) multiply top and bottom by sqrt(34) since square roots can't be in the denominator. =(3sqrt(34))/34.

find the exact value of the expression whenever it's defined

given: arccos[cscpi/2+sin(11pi)/6]. it's best to work from the inside out. the first part we can evaluate is cscpi/2 which is the reciprocal of sine. 1/sinpi/2=1. sinpi/2=1 because pi/2 is the same as 90º on a unit circle or quadrant which is (0,1) and since 1/sinpi/2=x/y y=1. now we can evaluate sin(11pi)/6. for this one we can reference the chart to see what sinpi/6 is and it's 1/2. however, we need to check what quadrant (11pi)/6 is in. by switching it to degrees (*180/pi) we can determine it's in the fourth quadrant. only cosine is positive in the fourth quadrant so our answer will be negative. we then have arccos(1+-1/2)=arccos(1/2). since arccos is the same as cos^-1, we can refer to the chart to determine what angle gives us 1/2 which is pi/3.

the vectors a and b represent two forces acting on the same point. if theta is the smallest positive angle between vectors a and b, approximate the magnitude of the resultant force.

in this problem, we are given vector a, vector b, and the angle in between them, known as theta. its very helpful to sketch this one out. the resultant force is what we are looking for, and its what happens when the two angles are combined. picture vector b as a river and the current flows in the direction of vector b. you are trying to swim the way vector a is going, so you will likely end up being moved by the current somewhere in between the two vectors. after drawing a line in between the two vectors, connect the remaining points with dotted lines, creating a parallelogram. the side we are trying to find is the line in the middle, which we will call "X". since the opposite sides of a parallelogram are the same, we can label the opposite of both vectors the same length. now it can be noted that there are two triangles. look at one of the triangles and draw it off to the side. in a parallelogram, two angles next to each other add up to 180º. subtract theta from 180º to get the angle of the other side. for these problems, we will always use law of cosines. use the formula: X^2 = a^2+b^2-(2)(a)(b)cos(angle across from side X). plug it into your calculator (do not round) then take the square root to get the final answer.

solve the triangle ABC given the conditions: a=11 b=18 and B=53º

it helps to visualize this one by drawing a picture. remember for the given side lengths, the side is across from the respective angle. for example, angle A is across from side a. for the given problem, we can't assume this is a right triangle so the Pythagorean theorem can't be used. here we can use the law of sines or the law of cosines. since we know both b's (angle B and side b) it's best to use the law of sines. for this example, we would do sin53/18 = sinA/11 in order to find the value of angle A. multiply both sides by 11 to get rid of the fraction. since sin is still attached to the A we need to take the inverse of the equation on the opposite side. sin^-1(11*(sin53)/18) which we plug into a scientific calculator (in degree mode) to get approximately 29.21º. we can now find C easily since all angles in a triangle add up to 180º, we can take 180-53-29.21 = 97.79º. now we only need side c which we can find the same way we found angle A. sin53/18 = sin97.79/c which can be cross multiplied to get csin53 = 18sin97.79 then divide both sides by sin53 to get c by itself and plug the equation into a calculator to get c = 22.33. to check that all your answers are correct, the smallest angle should be across from the smallest side, and the largest angle across from the largest side.

find all solutions of the equation on [0,2pi) given: sinx+sin2x=0

sin2x=2sinxcosx take out the GCF which is sinx. sinx(1+2cosx)=0. set both factors equal to zero. sinx=0 and 1+2cosx=0. sin doesn't equal zero in the chart so it's a quadrantal angle. sin is a y-value, so we draw the quadrants and figure out which points have a y-value of 0 and those are the two on the x-axis. the radian values are 0 and pi so those are the answers for sinx=0. if we solve for the other one we get cosx=-1/2 and we can use the chart for this one. the reference angle is pi/3 but we have a negative value so we need to think about what quadrant cosine is negative in. this is both quadrant 2 and 3. if the angle is in the first quadrant we use the chart, but if it's in the 2nd, we do one less than the chart (2pi/3). if it's in the third quadrant we do one more (4pi/3), and in the last quadrant we do double one less (5pi/3). so the answers for this one would be (2pi)/3 and (4pi)/3. in order, the final answers are 0, (2pi)/3, pi, and (4pi)/3.

find the exact arc length and area of the sector, given the following: r=10m and theta=60º

the formula for arc length is s=(r)theta. the formula for area of a sector is A=1/2(r^2)theta. in order to use either of these, theta MUST be in RADIANS. to convert degrees to radians multiply it by pi/180. for this problem, that gives us pi/3. we calculate the arc length to get (10pi)/3 meters (m). we calculate the area of the sector to get (50pi)/3 meters squared (m^2) since area is always squared.

insert two geometric means between a and b given a=2 and b=54

use the formula a_n=a_1(r)^(n-1) since there are two terms in between a and b, term a is a_1 and term b is a_4. the terms we are looking for are a_2 and a_3. the a_n in the formula can be used for any given number besides a_1 and its typically the last number in the sequence, so we will use b. remember that a_4 (which is b) isn't going to be the same number as n by itself. using the example, b=54 and its the fourth number in the sequence (a_4), so n=4 since a_n=a_4. solve for r. in this example, this leaves us with 27=r^3 so we need to take the cube root which you can easily do on a scientific calculator. ask your teacher if you don't know how. we then find that r=3, so multiply a_1 by three to get a_2 and a_2 by 3 to get a_3. in this example, a_2=6 and a_3=18.

find all solutions of the equation given: sin^2x+2cos^2x-2=0

when we answer these equations, there are technically infinately many solutions, so we add 2pin to our answer. this means we can go around a unit circle an infinite number of times to get all solutions. for this equation we want to have the same trig identities. sin^2x is the same as 1-cos^2x. now there are only cosines in the equation: 1-cos^2x+2cos^2x-2=0. after solving for cosx we get cosx=+-1 do not forget the +-. this isn't on the chart so it's a quadrantal angle. cosine is x-values so we need to use the angles with an x-value of 1 and those are 0 and pi. the final answers are 0+2pin and pi+2pin.