Unit Circle!

A full rotation around a circle is 360 degrees. In a the Unit circle only certain angles needs to be remembered.  0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, and 360°! How can All these certain angles be memorized!?!

=tep one: Draw a circle with a cross going through it. m =tep two: We have learned that a circle starts from 0 and ends at 360, so that can be put at start. m =tep three: The circle is cut into 4ths , so at the black dot is ¼ of the angle from the circle is used. ¼ of 360 is 90. At the blue dot is ½ of the angle from the circle is used. ½ of 360 is 180. AT the purple dot is ¾ of the angle from circle is used. ¾ of 360 is 270. m

=tep four: Lets Just use ¼ of the circle! 90 m =tep five: The red point is ½ of the angle of 90 is 45. The yellow point is 1/3 of the angle of 90 is 30. The green point is 2/3 of the angle of 90 is 60. m OK! Almost Done with angles! m

0

90

60

45 30

=tep =ix: Quadrant 2·s

angles are just reflections 180 0 of Quadrant 1, but plus 360
 90 (because it passes 90). 30+90=120 45+90=134 60+90=150 . The same with 270 Quadrant 3, but add 180. 30+180= 210 45+180=225 60+180=240. One more time with Quadrant 4 using 270! 30+270= 300 45+270=315 60+270=330. m THAT·= THE DEGREE=! NEXT«« m

No need to fret. The radians of the unit circle are equivalent to the degrees. m In a full rotation of a circle (in radians) equals 2 Ǒ. m Ǒ/6, Ǒ/4, Ǒ/3, Ǒ/2, 2 Ǒ/3, 3Ǒ/4, Ǒ, 7 Ǒ/6, 5 Ǒ/4, 4 Ǒ/3, 3 Ǒ/2, 5 Ǒ/3, 7 Ǒ/4, 11 Ǒ/6, 2 Ǒ. m How can ALL those radians be memorized !?! m

emorizing the Degrees is important in finding the radians that corresponds to that particular angle. m The formula for converting the lj° angle measurement is: lj° X(Ǒ/180) radian at lj° Ǒ/6 m Example: 30 ° X(Ǒ/180) m †ut«if you forget the formula or m the corresponding angles.. That·s ok m Just use the technique that was used for angles. m

=tep one: Draw a circle with a cross going through it. m =tep two: We have learned that a circle starts from 0 and ends at 2Ǒ, that can be put at start. m =tep three: The circle is cut into 4th , so at the black dot is ¼ of the radian from the circle is used. ¼ of 2Ǒ is Ǒ/2. At the blue dot is ½ of the radian from the circle is used. ½ of 2Ǒ is Ǒ. At the purple dot is ¾ of the radian from circle is used. ¾ of 2Ǒ is 2Ǒ/3. m

=tep four: Lets Just use ¼ of the circle! Ǒ/2 m =tep five: The red point is ½ of the radian of Ǒ/2 is Ǒ/2. The yellow point is 1/3 of the radian of Ǒ/2 is Ǒ/6. The green point is 2/3 of the radian of Ǒ/2 is Ǒ/3. m OK! Almost Done with Radians! m

0

Ǒ/2

Ǒ/3

Ǒ/4

Ǒ/6 =tep =ix: Quadrant 2·s

radians are just reflections 0 of Quadrant 1, but plus Ǒ 2Ǒ Ǒ/2(because it passes
 Ǒ/2 ). Ǒ/6+Ǒ/2=2 Ǒ/3 Ǒ/4+Ǒ/2=3 Ǒ/4 Ǒ/3+Ǒ/2=5Ǒ/6. The same with Quadrant 3, but 3Ǒ/2 add Ǒ. Ǒ/6+Ǒ =7 Ǒ/6 Ǒ/4+Ǒ/=5 Ǒ/4 Ǒ/3+Ǒ=4Ǒ/3. One more time with Quadrant 4 using3Ǒ/2! Ǒ/6+3Ǒ/2=5 Ǒ/3 Ǒ/4+3Ǒ/2=7 Ǒ/4 Ǒ/3 +3Ǒ/2=11Ǒ/6 . m THAT·= IT FOR RADIAN=! NEXT«.

m

The points might look a little intimidating at first, but applying a one unit radius circle to a graph makes it much easier. m (X,Y) coordinates are corresponding points t on the circumference of t the circle to the angles a and radians you just a learned. m HOW?!? m

=tep one: Draw a circle with a cross going through it. m =tep two: Imagine the circle (with a radius of one)on a graph with the midpoint at (0,0). m =tep three: That would make start (1,0), the black dot (0,1), the blue dot (-1,0), and the purple dot (0,-1). m

=tep four: Lets Just use ¼ of the circle! m =tep five: memorize the 0,1 numbers.(¥ 3/2) (1/2) (¥ 2/2) for these are the only other numbers you need. m =tep six: On the yellow dot the x coordinate is the longest out of all of them, and the largest number you memorized is (¥ 3/2). The y 1,0 coordinate is the shortest of all of them, and the smallest number you memorized is (1/2).The yellow dot·s full coordinate is ((¥ 3/2),(1/2) ). The red dot·s x and y coordinates are both in the middle, and the middle you memorized is (¥ 2/2). The red dot·s full coordinate is ((¥ 2/2), (¥ 2/2)). That also leaves you to figure that the green dot·s coordinate is ((1/2),(¥ 3/2)). m

(1/2),(¥ 3/2)

0,1

m

(¥ 2/2), (¥ 2/2)

=tep seven: Quadrant 2·s coordinates are just mirror

reflections of Quadrant 1, -1,0 1,0 with the exception that Q2 causes the X coordinates
 to become negative. (-(1/2),(¥ 3/2)) (-(¥ 2/2), (¥ 2/2)) (-(¥ 3/2),-(1/2)). Q3 coordinates 0,-1 reflect directly from Q2, with the exception that Q3 causes both X and Y coordinates to be negative. (-(1/2),-(¥ 3/2)) (-(¥ 2/2), -(¥ 2/2)) (-(¥ 3/2),-(1/2)). Q4 coordinates reflect directly from Q3, with the exception that Q4 cause only the Y coordinate to be negative. ((1/2),-(¥ 3/2)) ((¥ 2/2), (¥ 2/2)) ((¥ 3/2),-(1/2)).

(¥ 3/2),(1/2)

A


1 = radius Y Ê

 Ê   X  
 Ê 


Ê 

Ê



 Ê Ê Ê  ÊÊ
  Ê Ê Ê   tanlj =Y and Y= Y= sinlj and X=X=coslj Remember X 1 1 the radius equals one  cot lj= X Y Cotangent is just the Reciprocal Identity of Tangent which is next

1 = radius  X

csclj= 1 sinlj= 1 seclj= 1 sinlj csclj coslj coslj= 1 cotlj= 1 tanlj= 1 seclj tanlj cotlj Reciprocal Identities is the inverse of the sign.  sinlj= Y or Y. vÊ the sign to 1 = csclj = 1 Remember 1

Y

sinlj

coslj=X or X . vÊ the sign to 1 = seclj = 1 1

X

coslj

 tanlj= Y. vÊ the sign to X = cotlj= 1 X

Y

tanlj

the radius equals one

Y

1 = radius

sin²lj+cos²lj =1 1-tan²lj = sec²lj 1+cot²lj =csc ²lj  No need to fear! Interpret it to the Pythagorean theorem. x² + y² = r² sin²lj= x² and cos²lj= y². =o that fits directly in with the Pythagorean theorem x² + y² = r².

Y X

sin²lj+cos²lj =1². 1²= 1

1= 1²= r²

and tan²lj= O² . =o rearrange the Pythagorean theorem to ² r²- y² = r² . 1-tan²lj = sec²lj. sec²lj= r² = 1² = 1 x² x² x² x² x²

 1= 1²= r² and cot²lj= Ò² . Again rearrange the theorem to r²+ x² = 1+cot²lj =csc ²lj.

O² csc ²lj= r² = 1² = 1 O² O² O²

O²

r² O²

1 = radius  X

In Degrees sin(90° lj)=coslj cot(90° lj)=tanlj

cos(90° lj)=sinlj sec(90° lj)=csc lj

ǃ

tan(90° lj)=cotlj csc(90° lj)=seclj

A triangle has 180º. The right angle takes up half. While 90º remains. NOTICE THE CONNECTION OF THE C & O: The measures of the angles lj LETTERS * SINE AND COSINE CO-FUNCTIONS * TANGENT AND COTANGENT COand ǃ add to 90º.These angles FUNCTIONS * SECANT AND COSECANT COare complementary angles. FUNCTIONS * COMPLEMENTARY  lj+ǃ=90º{ ǃ=90º-lj} or {lj =90º-ǃ} sin lj=cosǃ{ sinlj-cos(90º-lj)} or{ cosǃ-sin(90º-ǃ)}

To make in to radians; just switch 90° with Ǒ/2.

Y

sin( lj)= sinlj cos( lj)=coslj tan( )= tanlj csc( lj)= csclj sec( lj)=seclj cot( lj)= cotlj The easiest way to remember this is: If the angle is negative then so is the sign If the angle is positive then so is the sign ««..but does not apply to cosine and secant

>

>

sin(lj+ǃ)=sinljcosǃ+cosljsinǃ sin(lj ïǃ)=sinljcosǃïcoslj sinǃ cos(lj+ǃ)=cosljcosǃïsinljsinǃ cos(ljïǃ)=cosljcosǃ+sinljsinǃ Notice: In the sine formulas, + or ï on the left is also + or ï on the right. =ine also has a pattern of sin cos cos sin. †ut in the cosine formulas, + on the left becomes ï on the right. Cosines pattern is cos cos sin sin and; and vice-versa.

tan(lj+ǃ)=tan lj+tanǃ 1-tanljtanǃ

tan(lj-ǃ)=tan lj-tanǃ 1+tanljtanǃ

Notice: In the tangent formulas, + or on the left is also on the right top, while the bottom right is not. Also that tangent is the only sign used.

sin2A=2sinAcosA cos2A=1-2sin²A tan2A= 2tanA 1-tan²A

cos2A=cos²A sin²A cos2A=2cos²A-1


Ê Ê 
 
m
 m

  
  
  m  
Ê Ê  
Ê 
 

 
Ê Ê  sin2A=sin(A+A). Use the sine of sums identity sinAcosA+cosAsinA Which equals 2sinAcosA.

Half angle identities come from the double angle identities. sin²A=½(1-cos2A) cos²A=½(1+cos2A) tan²A=½(1-cos2A) (1+cos2A) take the double angle formula you memorized (cos2A=12sin²A) change the 2A=a, so it changes to a/2. cosa=1ï2sin2 (ǂ/2) Leave only sin2 (ǂ/2) to solve for. Giving us sin²A=½(1-cos2A). Taking the same steps for cosine and tangent as well.

These or for oblique triangles!! (not right triangles) † a = b = c sinA sin† sinC c

a

h

When solving for oblique A b Triangles, at least one side is needed and two other parts that can be either sides or angles. PROOF-LAW OF =INE= sinA= h/c ; h=csinA asinC = csinA =inc+h/a; h=asinC sinAsinC sinAsinC asinC=csinA a = c sinA sinC

C

Any two angles and any side! m =AA or AA= or A=A m

Given: sinA=60° sin C= 75° b= ¥ 2 m<†= 180 °-(60 °+ 75 °)=45 ° a = b a = ¥2 sinA sin† sin60 sin45 ¥3 = c a = c sinA sinC sin60 sin75

† c

a

A b

a = ¥2 a= ¥ 3 ¥ 2/3 ¥ 3/2 ¥3= c c=3.549 ¥ 2/3 .9659

C

Any two angles and any side! m =AA or AA= or A=A m

Given: sinA=58° sin †= 64° c= 12 m
† c

a

A b

a = 12 .848 .848 12 = b .848 .899

a= 12 b=12.7

C

If b) (0<sin†<1) 0: (a
andb) (0<sin†<1) 2: (a1)

An angle and two sides! m ==A m

† c

a

Given: sinA=55.6° A A=8.44cm b= 25.1cm a = b 8.44 = 25.1 sinA sin† sin 55.6 sin†

8.44 = 25.1 .825 sin†

sin†= 2.45 sin-1 (2.45)= ERROR

NO TRIANGLE!!

b

C

An angle and two sides! m ==A m

† c

a Given: sinA=43.5° A C a=10.7cm c= 7.3cm b a = C 10.7 = 7.3 10.7 = 7.3 =inC= 27.9 sinA sinC sin43.5 sinC .688 =inC 180-27.9=152.4 152.4+43.5= OVER 180€ (Not possible for 2nd triangle) Only one m<†= 180 °-(43.5 °+ 27.5°)= 109 a = b 10.7 = b 10.7 = b b=14.7cm sinA sin† sin43.5 sin109 .688 .946

The square of a side is Equal to the sum of the square of the other two sides minus two times the product of those two side and the cosine of the inclined angle. a²=b²+c²-2bc(cosA) b²=a²+c²-2ac(cos†) c²=a²+b²-2ab(cosC) These are also for oblique triangles! Needs any two sides but and angle that does not correspond to the sides.

An angle and two sides! Or All three sides † m === or =A= Find sidea c Given: cosA=42.3° A b b=15.4cm c= 12.9cm a²=b²+c²-2bc(cosA) a²=15.4²+12.9²-2(15.4)(12.9)(cos42.3) a²=125.3cm a=10.5cm m

a

C

An angle and two sides! Or All three sides † m === or =A= Find A c Given: a=37.6 b=42.9 c= 62.7 A b a²=b²+c²-2bc(cosA) Rewrite: -a²+b²+c² = cosA 2bc -62.7²+37.6²+42.9² = cosA 2(37.6)(42.9) cosA= .2099 cos-1 (.2099)= 77.89 m

a

C

Radian measure of lj = m

s



 -The circumference of a circle is r an arc length. -The ratio of the circumference to the diameter is the basis of radian measure. (why it is not in degrees!!) Theorem~ In any circle the same ratio of arc length to radius will have the same central angle measurements. Proportions!~ s1 = s2 ONLY IF!! lj1= lj2 r1 r2

s

A person moves a wheel that·s radius is 10 meters, and found herself 8 meters from where she began. What is the central angle (degrees) the wheel has moved? lj = s = 8 = .8 (radians answer) r 10 (degrees)lj = 180(radians) lj = 180(.8) Ǒ Ǒ (degrees) lj= 144/ Ǒ lj=45.84°

 r

s

There are two gears touching one another. One has a radius of 10cm and has rotated 8cm around. What is the rotation and the central angle of the other gear if the radius is 5cm? lj = s = 8 = .8 (radians answer) r 10 (degrees)lj = 180(radians) lj = 180(.8) Ǒ Ǒ (degrees) lj= 144/ Ǒ lj=45.84 ° (The gears a porpotional!) .8= 8 = s2 s2= 4cm lj = s1 = s2 r1 r2 10 5

 r

s  r

The area of a sector of a circle can be calculated by degrees or radians.

(radians)

lj (r²)= Area 2

(Degrees)

lj (Ǒ)(r²)= Area 360

A  r

Theorem~ The area of the circle is proportional to 360Ý Area of sector = Angle of sector Area of circle 360 or (2Ǒ)

rs. Parker ordered a large chesse pizza. The radius of the pizza was 5 ft. she made a bet with that class she could eat a slice of pizza with a central Angle of 60 degrees. How much pizza Did she consume? Area of sector = Angle of sector Area of circle 360 Area of sector = 60 Ǒ5² 360 Area of sector = 60 (25Ǒ) That·s a lot of 360 pizza. Area of sector= 13.09ft²

A  r

rs. Parker baked a circlar brownie. Her neighbor wanted 35.4cm² of the brownie pie. About what angle(radians) did rs. Parker cut her brownie, if the radius was 6cm. Area of sector = Angle of sector Area of circle 2Ǒ 35.4 = Angle of sector Ǒ6² 2Ǒ Angle of sector = 35.4 (2) 36 Angle of sector= 1.967(radians)

A  r