T formulae questions (4 Viewers)

[clintmyster]

i got these questions for hw and kinda stuck how to do them..I know the three equations for t for sin cos and tan but im a bit stuck how to use them..solutions would be much appreciated!

 

[lyounamu]

i got these questions for hw and kinda stuck how to do them..I know the three equations for t for sin cos and tan but im a bit stuck how to use them..solutions would be much appreciated!

1) y^2 = 8x
x = y^2/8
dx/dy = y/4
dy/dx = 4/y (i.e. the gradient of a tangent)

However, the tangent is parallel to the line y = 2x+3, which means that it will have a gradient of 2.

Therefore, dy/dx = 4/y = 2
Therefore, y = 2
Therefore, x = 1/2 (found this by substitution).

2) y^2 = 12x
x = y^2/12
dx/dy = y/6
dy/dx = 6/y (gradient of a tangent)
dy/dx = - y/6 (gradient of a normal)

However, this normal is parallel to the line 2x + 3y + 5 = 0 (which is basically equal to y = -2x/3 - 5/2).
So the normal's gradient is -2/3

Therefore, -y/6 = -2/3
y = 4
Then, x = 4/3

 

[clintmyster]

I feel so stupid, I uploaded the wrong screenshot....below are the questions i need help with

 

[lyounamu]

I feel so stupid, I uploaded the wrong screenshot....below are the questions i need help with

Are you serious? I knew that something was wrong...

Ok. Let me try again then.

 

[clintmyster]

yeah im dead serious! I actually asked the questions before and entitled it 'questions' but it seems that the one i called 'questions' this time was in a different folder and that was why this came up!

Thanks!

 

[lyounamu]

yeah im dead serious! I actually asked the questions before and entitled it 'questions' but it seems that the one i called 'questions' this time was in a different folder and that was why this came up!

Thanks!

I will just do (f) as all the questions above are similar and what you did to do is to substitute the appropriate value of tan by making t = tan 1/2 x (x is the value that you are trying to find).

f) (2tan(11pi/12))/(1-tan^2(11pi/12)) = tan (11pi/6)
= tan 11pi/3
= tan 330
= -1/square root of 3

 

[clintmyster]

Thanks for the solution. I dont quite understand how this step works

2tan(11pi/12))/(1-tan^2(11pi/12)) = tan (2 . 11pi/6)

also in class today we learnt some substitutions that replaced tan, cos and sin
can be seen here: http://en.wikipedia.org/wiki/Tangent_half-angle_formula#Identities

I was wondering how do you use those to solve the equation?

 

[lyounamu]

Thanks for the solution. I dont quite understand how this step works

2tan(11pi/12))/(1-tan^2(11pi/12)) = tan (2 . 11pi/6)

also in class today we learnt some substitutions that replaced tan, cos and sin
can be seen here: http://en.wikipedia.org/wiki/Tangent_half-angle_formula#Identities

I was wondering how do you use those to solve the equation?

I am sorry. I didn't explain that. I used the double angle formula there.

tan 2x = 2tanx/(1-tan^2(x))

I actually should have used t-formula. But it is dragging me down when I write the one up here. It is just a tedious work.

Just remember, when you substitue the t-formula, always let t = tanx/2 (where x is the one that you are trying to find from the question).

So when you get the question like: "find tan2x in terms of t", you let t = tanx rather than t = tanx/2.

 

[clintmyster]

its just that the question wants me to solve in terms of t and i dont really know how for this question..

i noticed the double angle formula there but then shouldnt:

2tan(11pi/12))/(1-tan^2(11pi/12)) = tan (11pi/6) ?

did you like jump a step and then equate that to t? so then you multiply by 2?

 

[lyounamu]

its just that the question wants me to solve in terms of t and i dont really know how for this question..

i noticed the double angle formula there but then shouldnt:

2tan(11pi/12))/(1-tan^2(11pi/12)) = tan (11pi/6) ?

Hahahaha... I made a mistake there. Yeah, you are right. I wrote straight on the computer so I probably miswrote it.

 

[clintmyster]

Hahahaha... I made a mistake there. Yeah, you are right. I wrote straight on the computer so I probably miswrote it.

you shoulda just said you were testing me!

with 4b) tho you cant use the double angle formula for tan. Or can you?

 

[lyounamu]

you shoulda just said you were testing me!

with 4b) tho you cant use the double angle formula for tan. Or can you?

You cannot. So you have to make t = tan7.5 degrees
and substitute tan 15 = 2t/1-t^2 for every singe tan 15. It's a hard-work. I have to blame my laziness on this.

 

[clintmyster]

You cannot. So you have to make t = tan7.5 degrees
and substitute tan 15 = 2t/1-t^2 for every singe tan 15. It's a hard-work. I have to blame my laziness on this.

and when you do this you get answer in terms of t. Does that mean you replace t again with tan7.5? can you list all the steps for me to get the answer? i dont quite get how you can end up with an exact value out of this especially with 15 unless you use 45 and 30 and subtract? As you can see im quite lost!

 

[lyounamu]

and when you do this you get answer in terms of t. Does that mean you replace t again with tan7.5? can you list all the steps for me to get the answer? i dont quite get how you can end up with an exact value out of this especially with 15 unless you use 45 and 30 and subtract? As you can see im quite lost!

1. Let t = tan 7.5
2. Therefore, tan 15 = 2t/(1-t^2)
3. Substitute that in
4. Factorise
5. Factorise
6. Look at whatever you get.

If you are really lost, I will come back and give you the full solution. I wish someone can fill in my laziness.

 

[clintmyster]

you be the judge LOL..it seems way too complicated then it should be! can you verify those steps are right. And if they are do I sub tan 7.5 back in?



be warned its really big!

 

[lyounamu]

you be the judge LOL..it seems way too complicated then it should be!



be warned its really big!

That's my answer as well.

 

[clintmyster]

iight now what?

the answer is sin 30 = 1/2

and that completely entirely (add some other adjective here) puzzles me!

 

[lolokay]

you shoulda just said you were testing me!

with 4b) tho you cant use the double angle formula for tan. Or can you?

from the t formulas, you should know that sin@ = 2t/(1 + t^2)

 

[clintmyster]

from the t formulas, you should know that sin@ = 2t/(1 + t^2)

i know that but how do you use that in this scenario..i dont see many ways of making this in terms of sin unless you convert tan...is this question supposed to be hard? its supposedly the "easier" questions in yr12 3u!

 

[lyounamu]

iight now what?

the answer is sin 30 = 1/2

and that completely entirely (add some other adjective here) puzzles me!

Now you put that back in and factorise. You should get that.