skeTching triG grAphs quEstion (1 Viewer)

[+GriM ReApeR+]

sketch for 0 <= x <= 2pi

a) y = 4sin(2x- pi/2)

b) y = cos(2x - pi/4)

okok....i know the amplitudes, periods and phase shifts for each one, but every time i sketch them , i get the wrong answer*, so can somebody plz tell me how to sketch these #$^#@ graphs step by step...

Thanks

* i am definately sure that the textbook answers are correct as i also checked them on a graphic calculator


EDIT: sorri pplz...a) was actually y = 4sin(2x + pi/2)...oops...^^"

 

[:: ck ::]

a) y = 4sin(2x- pi/2)

first note the amplitude = 4, Period = 2pi/2 = pi

... then shift the whole graph of y=4sin2x by pi/2 to the right, such that when x = 0, y = -1 ...

b) y = cos(2x - pi/4)
amplitude = 1, period = 2pi/2 = pi

shift the whole graph of y=cos2x by pi/4 to the right.. [ a tiny bit more tricky ]... so when x = 0, y = 1/sqrt2 ... almost at 1 ...


usually i find it easiest to draw the curve of cos2x or 4sin2x in pen... but axis in pencil... then readjust ur graph axis after

if u dont know where to start... sub values of x in for pi/2, pi etc...

 

[AGB]

speaking of trig graphs... how do u graph tan(x + y) = 1 ?? it was in my half yearly and i didnt get it

 

[nike33]

isnt it tan(x+y) = 1 x+y = pi/4 which is a straight line?

 

[:: ck ::]

tan(x+y) = 1
x+y = tan^-11
x + y = pi/4
y = pi/4 -x ?

is that it ?

editz : damn nike beat me

 

[+GriM ReApeR+]

how come i am still getting the wrong answer....T_T....
i get something that looks like a negative sine graph with amplitute 4 and period of 2pi...because the answer looks like the graph has been shifted by pi/4...???

 

[Grey Council]

lol @ AGB.

(thats payback for picking on me in that PDHPE forum )

and how I sketch them is this:
sketch it using pencil.
rub out axis.
put in new axis once shifted. And I don't know what you are doing wrong. You know amplitude, phases and periods for both..
i think your phases must be wrong. i think

EDIT: hehe, check next post. Ryan actually did the graph. ^_^

 

[:: ck ::]

Originally posted by +GriM ReApeR+
how come i am still getting the wrong answer....T_T....
i get something that looks like a negative sine graph with amplitute 4 and period of 2pi...because the answer looks like the graph has been shifted by pi/4...???

the period should be pi

the coefficient of x is 2.. so period = 2pi/2 = pi

 

[+GriM ReApeR+]

so...

step 1: draw 4sin2x
step 2: shift left by pi/2

is this method correct??... i keep getting it wrong!!argh!
I must be doing something wrong...

 

[:: ck ::]

shift right!

 

[+GriM ReApeR+]

Originally posted by :: ryan.cck ::
shift right!

y = 4sin(2x + pi/2)
shift right? not left?

 

[CM_Tutor]

AGB - sketching tan(x + y) = 1 is pretty hard for Extn 1, but the answer is a series of parallel lines, each of
gradient -1, with y-intercepts at (4n + 1) * pi / 4, where n is an integer

 

[:: ck ::]

mmm how does it become a series of parallel lines?

oh yeh ... crap.. general sol

 

[nike33]

ahh yes..but if they give a domain

 

[nike33]

cause arctan 1 = (pi) n + pi/4

 

[:: ck ::]

y = 4sin(2x + pi/2)
shift right? not left?

ur original question was y = 4sin(2x - pi/2)

 

[CM_Tutor]

Originally posted by nike33
ahh yes..but if they give a domain

True, but the whole point of such a question is NOT to give a domain. As it stands, the trap is for people to take inverse tan of both sides and get x + y = pi / 4, and plot a single line. There is little point in asking the question if you take out the trick.

 

[+GriM ReApeR+]

Originally posted by :: ryan.cck ::
ur original question was y = 4sin(2x - pi/2)

look at the edit note

 

[AGB]

Originally posted by CM_Tutor
AGB - sketching tan(x + y) = 1 is pretty hard for Extn 1, but the answer is a series of parallel lines, each of
gradient -1, with y-intercepts at (4n + 1) * pi / 4, where n is an integer

oh i get it now

yeh i think it was in the ext 2 half yearly...cant remember tho...