2pi/ absoulute B

where be is only going to be a positve integer

phase shifting moves the whole graph either left or right of (C)

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A positive phase shift is a shift to the right

find the amplitude period phase shift and full plot for (m=4 n=6)

[image]

amp= n

period = one full cycle = 2mn

phase shift can be determined by logic- if originally sin graph starts at 0 and moves to one, this graph starts at o to -1 (INVERSED) so we know a shift occurred

+ ps.= moves to R

-PS= moves to the Left

PS= -4n bec it moved one complete peak

-to write out formula work backwards: if period

8pi=pi/b solve for b plug into

if phase shift =

-c/b=-4n ,solve for c etc.

questions: why is phase shift n as answer and pos in formula???

while doing verification on 6.1 if you see one side with the numerator in the format

(1+c¹) or (1-c¹) use the foil of trigometric function

c²+s² =1

but take the invere sign of the problem

if you see a problem with cubes solve

sin³ + cos³ what do you do?

law of cubes:

foil

(sin¹ +cos¹ ) (sin² - sincos+cos² )

take the opp of the sign on first part

sin2 + cos2 =1 ; solve from there

cofunction means using the formula

sinO=Cos (pi/2 -O)

remember completementary comes before supplementary

C comes before S

90 comes before 180

for tan(x-pi/2) or tan(O-pi/2) inorder to verify tangent if original formula

(tan(u-v) =tanu -tanv / 1+ tanutanv) does not work

what can you do?

tan(u-v) = sin(u-v) /cos (u-v)

solve.

move right units over to left tran form into sin(4t-t)=  o sin's orgin is at pi so...

3t= pin

t= pi/3 so pi in intervals of 0-pi would be your answer

3 double angle formulas for cos2O

special property to depend what quarant cos tan sin is in when using double angle??

cos 2α = cos2 α − sin2 α

cos² α − sin² α

= (1− sin² α) − sin² α

= 1− 2sin² α

 

cos² α − sin² α

= cos² α − (1 − cos² α)

= 2cos² α − 1

 

multiply given angle by 2 inorder to determine positive or negative quadrant

cos 2α = cos2 α − sin2 α

 

cos 2α = 1− 2sin2 α

 

cos 2α = 2cos2 α − 1

sin(α + β) = sin(α + α) = sin 2α

Consider the RHS:

sin α cos β + cos α sin β

If we replace β with α, we obtain:

sin α cos α + cos α sin α = 2 (sin α cos α)

basically for the double-angle formulas you are taking the previous formulas like cos(alpha +beta)= cosalphacosBeta -sinalphasin beta and making them all one angle alpha +alpha

tangent is the same way

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half angle formulas

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replace x with x/2 for half angle identities

for half angle formulas if given an alpha plug diretly into equation ignore sin or cos in respect to sin2 theta solve it as

(theta)2 NOT sin ^2 (theta)

when you have a term of cos/sin 4u know that in the conversion to double angle formula u in formula will = 2u because

sin 4u= sin (2* 2u)

ex.

cos4u = cos (2* 2u) = 1-2sin^2 2u

by proving that RS does not equal LS

chosing quadrant with different signs

6.2 #9

find all solutions for

cos O = 1/secO

all values for which equation is defined except

O= pi/2 +pin

6.2 # 11

2cos 2theta - Sqrt # = 0

find all solutions

use unit circle, whenever you find all the solutions = them up to 2theta and solve for theta!!!

2theta = xpi/x

theta= xpi/2x

remember if you have a square root in finding solutions you have to solve for all positive and negative answers!!!***

6.2