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# Trig identities watch

1. i need help with b)ii) if someone could do the working it would be really helpful being trying at it for an hour and stil cant figure it out. i checked the answers also but no clue.
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2. (Original post by djmans)
i need help with b)ii) if someone could do the working it would be really helpful being trying at it for an hour and stil cant figure it out. i checked the answers also but no clue.
It's inverse cos of 0 first isn't it?
so
3. (Original post by will'o'wisp)
It's inverse cos of 0 first isn't it?
so
thats easy but how do you do b)ii) finding p and q
4. (Original post by djmans)
thats easy but how do you do b)ii) finding p and q
Oh right u didn't specify, nvm

So this is just graph transformations isn't it?

well we'll do q first because it's easier

so we know with that a cos graph it touches the x axis at 90°

A is the first bit of the graph we see from the diagram which touches the x axis at 100 degrees

There's a difference of 10 between them

Remember if we have y=f(x-a) we move the graph a units to the right

so q then is simply just 10
5. (Original post by will'o'wisp)
Oh right u didn't specify, nvm

So this is just graph transformations isn't it?

well we'll do q first because it's easier

so we know with that a cos graph it touches the x axis at 90°

A is the first bit of the graph we see from the diagram which touches the x axis at 100 degrees

There's a difference of 10 between them

Remember if we have y=f(x-a) we move the graph a units to the right

so q then is simply just 10
6. (Original post by djmans)
thats easy but how do you do b)ii) finding p and q
p is a bit more difficult

so we know the difference between where a cos graph normally touches the x axis is at the points 90° and 270° between the graph drawn in the range 0 to 360

the difference between A and B is 120

yet the original cos graph has a difference of 180 so to get the ratio between the given graph and the cos graph we do 120/180 which is 2/3

remember y=f(2x) we divide the x value by 2 not multiply the x value by 2 so in a sense we do 1/2

so for our 2/3 instead of 2/3 we do

which is of course 3/2
7. (Original post by will'o'wisp)
p is a bit more difficult

so we know the difference between where a cos graph normally touches the x axis is at the points 90° and 270° between the graph drawn in the range 0 to 360

the difference between A and B is 120

yet the original cos graph has a difference of 180 so to get the ratio between the given graph and the cos graph we do 120/180 which is 2/3

remember y=f(2x) we divide the x value by 2 not multiply the x value by 2 so in a sense we do 1/2

so for our 2/3 instead of 2/3 we do

which is of course 3/2
p is right but q is supposed to 60
8. (Original post by djmans)
(Original post by djmans)
p is right but q is supposed to 60
hmmm, but if we move the cos graph 60° to the right then shouldn't A be 150° instead of 100°

plus i can see that from the original cos graph A is 10° to big and B is 50° too little, then again why would you want to add them
9. (Original post by djmans)

from considering cosine's periodicity and the interval in which A and B are in.

A different way to think about it would be to take and you know that this crosses at x=90 and x=270
Now do the mapping and you have and then just substitute the appropriate angles in.
10. (Original post by will'o'wisp)
hmmm, but if we move the cos graph 60° to the right then shouldn't A be 150° instead of 100°

plus i can see that from the original cos graph A is 10° to big and B is 50° too little, then again why would you want to add them
hmm could it be the answer in the book is wrong after all this question is from the text book.
11. (Original post by RDKGames)

from considering cosine's periodicity and the interval in which A and B are in.
so we supposed to add the differences together? o.o
12. (Original post by will'o'wisp)
hmmm, but if we move the cos graph 60° to the right then shouldn't A be 150° instead of 100°

plus i can see that from the original cos graph A is 10° to big and B is 50° too little, then again why would you want to add them
The problem with your solution is that you're trying to work out q without taking how stretched the graph is into account. RDKGames's solution avoids this problem.
13. (Original post by djmans)
hmm could it be the answer in the book is wrong after all this question is from the text book.
sorry, since i forgot to take into account stretchiness of graph, multiply 100° by 1.5 which is p then we can see the difference because multiplying by 1.5 removes the stretchiness and makes the graph "normal", we then see A is 150° and the difference between 90 and 150 is 60
(Original post by TimGB)
The problem with your solution is that you're trying to work out q without taking how stretched the graph is into account. RDKGames's solution avoids this problem.

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