I'm Reviewing 10.1
we weren't allowed to use calculators and we had to know the trig chart . . .
In order to do this you must know these formulas:
cos(alpha +/- beta)= cos(alpha)cos(beta) -/+ sin(alpha)sin(beta)
sin(alpha +/- beta)= sin(alpha)cos(beta) +/- cos(alpha)sin(beta)
Alpha and beta will come from the trig chart. You will either add or subtract them to get the angle needed.
Ex: find the exact value of sin15 degrees.
alpha= 45
beta= 30
45-30 = 15
sin(45-30)= (sin 45)(cos 30) - (cos 45)(sin 30)
=(square root of 2/2)(square root of 3/2) - (square root of 2/2)(1/2)
square root of 6/4 - square root of 2/4 = square root of 6 - square root of 2/4
sin 15= square root of 6 - square root of 2/4
Thursday, December 30, 2010
Malorie's Holiday blog #3
I'm reviewing 9.1 right triangles
We learned anout right angles and SOHCAHTOA.
In order to do this, you must know the following formulas:
Sin = opp/hyp
Cos= adj/hyp
tan= opp/adj
cot= adj/opp
sec= hyp/adj
csc= adj/hyp
(*this is only used for right triangles)
Example:
In triangle ABC, Angle A = 90 degrees, Angle B= 30 degrees, and a = 15. Find b and c.
in order to find c you must do cos 30=c/15
c=12.99
in order to find b, you must use the pythagorean theorem.
b= 7.5
We learned anout right angles and SOHCAHTOA.
In order to do this, you must know the following formulas:
Sin = opp/hyp
Cos= adj/hyp
tan= opp/adj
cot= adj/opp
sec= hyp/adj
csc= adj/hyp
(*this is only used for right triangles)
Example:
In triangle ABC, Angle A = 90 degrees, Angle B= 30 degrees, and a = 15. Find b and c.
in order to find c you must do cos 30=c/15
c=12.99
in order to find b, you must use the pythagorean theorem.
b= 7.5
Malorie's Holiday Blog #2
I'm reviewing 8.1 because it was easy.
When solving for theta, you get the trig function by itself and then take the inverse.
Ex: cos(theta)= 36
(theta)=cos-1(36)
Inverses have 2 answers with a few exceptions. Find where the angle is based on the trig function and if the number is tve or -ve.
Steps:
Take the inverse of tve # to find the quadrant one angle.
To find quad 2, ve must be negative and add 180.
To find quad 3, add 180
To find qaud 4, make it negative and add 360.
Example:
Solve for 0 degrees is less than or equal to theta less than 360 degrees.
cos(theta)= square root of 3/2
(on the trig chart)
(Theta)=30
cos is only positive in the first and fourth quadrant, we already know the first so now we need the fourth.
-30 + 360 = 330
(theta)=330
Final answer is (theta)= 30 and 330
When solving for theta, you get the trig function by itself and then take the inverse.
Ex: cos(theta)= 36
(theta)=cos-1(36)
Inverses have 2 answers with a few exceptions. Find where the angle is based on the trig function and if the number is tve or -ve.
Steps:
Take the inverse of tve # to find the quadrant one angle.
To find quad 2, ve must be negative and add 180.
To find quad 3, add 180
To find qaud 4, make it negative and add 360.
Example:
Solve for 0 degrees is less than or equal to theta less than 360 degrees.
cos(theta)= square root of 3/2
(on the trig chart)
(Theta)=30
cos is only positive in the first and fourth quadrant, we already know the first so now we need the fourth.
-30 + 360 = 330
(theta)=330
Final answer is (theta)= 30 and 330
Malorie's Holiday BLog #1
In this blog i'm looking back to the easiest stuff I learned. 7-4 reference angles.
To find a reference angle for sin and cos, you must follow the following:
Step 1: Figure out which quadrant the angle is in.
(if given an angle with pi, use pi as 180 and simplify)
Step 2: determine whether the function is positive or negative.
Step 3: subtract 180 from the angle measure until it is between 0 and 90 degrees.
Step 4: If the angle comes out to be 0, 30, 45, 60, or 90 degrees, use the trig chart to simplify. If not, then leave it as is or plug it in the calculator.
Example:
Find a reference angle for sin 6pi/5
sin (6x180)/5 = 1080/5 = 216
-sin216
this angle would be in the III quadrant because it is between 180 and 270. It would be negative because sin follows the y axsis and y is negative in this quadrant.
216-180 = 36
subtract from 180
your final answer would be -sin36 degrees.
To find a reference angle for sin and cos, you must follow the following:
Step 1: Figure out which quadrant the angle is in.
(if given an angle with pi, use pi as 180 and simplify)
Step 2: determine whether the function is positive or negative.
Step 3: subtract 180 from the angle measure until it is between 0 and 90 degrees.
Step 4: If the angle comes out to be 0, 30, 45, 60, or 90 degrees, use the trig chart to simplify. If not, then leave it as is or plug it in the calculator.
Example:
Find a reference angle for sin 6pi/5
sin (6x180)/5 = 1080/5 = 216
-sin216
this angle would be in the III quadrant because it is between 180 and 270. It would be negative because sin follows the y axsis and y is negative in this quadrant.
216-180 = 36
subtract from 180
your final answer would be -sin36 degrees.
Charlie's Christmas Holiday's Blog #4
one recent thing we did was arithmetic sequences. where you use the formula to find the certain number of the sequence that you want to find or just find the real formula to find that certain number of the sequence. the formula you use is t*n = t*1 + (n - 1) d; where t*n is the sequence anser you want to find, t*1 is the firse numbe of the sequence, and d is the number being added.
For example:
6, 12, 18, 24
t*n = 6 + (n - 1) 6
= 6 + 6n - 6
= 6n
~~>all you do is find what is being added, then plug into the formula.
For example:
t*1 = 3 and t*2 = 7, find t*6
t*n = 3 + (n-1) 4
= 3 + 4n - 4
= -1 + 4n
t*6 = -1 + 4(6)
= -1 + 24
= 23
~~> you find what is being added between the first 2, then plug into the formula. then change the t*n to t*6 and replace the n in the new formula with 6.
For example:
6, 12, 18, 24
t*n = 6 + (n - 1) 6
= 6 + 6n - 6
= 6n
~~>all you do is find what is being added, then plug into the formula.
For example:
t*1 = 3 and t*2 = 7, find t*6
t*n = 3 + (n-1) 4
= 3 + 4n - 4
= -1 + 4n
t*6 = -1 + 4(6)
= -1 + 24
= 23
~~> you find what is being added between the first 2, then plug into the formula. then change the t*n to t*6 and replace the n in the new formula with 6.
Charlie's Christmas Holiday's Blog #3
this first semester we also did law of sines. the formula (sinA/opposite leg = sinB/opposite leg) is used when you have a nonright triangle and have atleast one pair of an angle and oposite leg with either a different leg or another angle. to work this you just plug into the formula and cross mulitply like we use to do in like prealgebra back in the gap.
For example:
angle Z = 69
angle Y = 22
side y = 4
side z = ?
sin69/z = sin22/4
4sin69 = z sin22
4sin69/sin22 = z
z = 9.97
~~> you're suppose to draw a triangle for this too, but i'm not that talented to do it on here. for this you take angle Y and put it over side y and set it equal to angle Z over side z, then cross mulitply to solve for z. so side z equals 9.97.
For example:
angle Z = 69
angle Y = 22
side y = 4
side z = ?
sin69/z = sin22/4
4sin69 = z sin22
4sin69/sin22 = z
z = 9.97
~~> you're suppose to draw a triangle for this too, but i'm not that talented to do it on here. for this you take angle Y and put it over side y and set it equal to angle Z over side z, then cross mulitply to solve for z. so side z equals 9.97.
Charlie's Christmas Holiday's Blog #2
in another chapter we went through sine, cosine, and tangent for right triangles. for this we use the 'word' SOHCAHTOA. this means that sine equals opposite over hypotenuse, cosine equals adjacent oer hypotenuse, and tangent equals opposite over adjacent. to do this you are given the information that the triangle is a right triangle and either a side length and an angle other than the 90 degree one or two side lengths. in order to solve, though, a triangle must be drawn.
For example:
angle C = 90 degrees
angle A = 12 degrees
side b = 4
side a = ?
tan 12 = a/4
a = 4tan12
a = .9
~~> obviously i can't draw a triangle on here. but i took the opposite side length of angle A which was unknow and put it over the adjacent side length which was 4. since tangent is opposite over adjacent.
For example:
angle C = 90 degrees
angle A = 12 degrees
side b = 4
side a = ?
tan 12 = a/4
a = 4tan12
a = .9
~~> obviously i can't draw a triangle on here. but i took the opposite side length of angle A which was unknow and put it over the adjacent side length which was 4. since tangent is opposite over adjacent.
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