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TRIGO

I'm not understanding why tanθ tan \theta =724 = -\frac {7}{24}

Find the value Cos θ\theta and tan θ\theta, given that Sin θ\theta = -\frac {7}{25} and 270o<θ<360o 270^o < \theta < 360^o

https://gyazo.com/3cdeeda4af99568f00a1f8195d877884

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Reply 1
Original post by Naruke
I'm not understanding why tanθ tan \theta =724 = -\frac {7}{24}

Find the value Cos θ\theta and tan θ\theta, given that Sin θ\theta = -\frac {7}{25}
and 270o<θ<360o 270^o < \theta < 360^o

https://gyazo.com/3cdeeda4af99568f00a1f8195d877884

cosθ\cos \theta is positive in the fourth quadrant (think CAST diagrams) but tan is negative there.

In either case, knowing that cosθ\cos \theta is positive is enough. Because tanθ=sinθcosθ=negativepositive=negative\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\text{negative}}{\text{positive}} = \text{negative}.
Reply 2
Draw a right angle triangle, the sides will be length 7, 24 and 25. tanθ \tan \theta in the range 270θ360 270^{\circ} \leqslant \theta \leqslant 360^{\circ} is negative, draw a quick sketch of the tan curve if you don't understand this.
Original post by Naruke
I'm not understanding why tanθ tan \theta =724 = -\frac {7}{24}

Find the value Cos θ\theta and tan θ\theta, given that Sin θ\theta = -\frac {7}{25}
and 270o<θ<360o 270^o < \theta < 360^o

https://gyazo.com/3cdeeda4af99568f00a1f8195d877884

sin x = -7/25 -> sin^2x=49/625 -> cos^2x=576/625 -> cosx=+/-24/25. But 270<x<360, so by considering the cosine graph, cos is positive in this region, so cosx=24/25. Now tanx=sinx/cosx = (-7/25)/(24/25)=-7/24.
Reply 4
too late
Reply 5
Original post by Zacken
cosθ\cos \theta is positive in the fourth quadrant (think CAST diagrams) but tan is negative there.

In either case, knowing that cosθ\cos \theta is positive is enough. Because tanθ=sinθcosθ=negativepositive=negative\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\text{negative}}{\text{positive}} = \text{negative}.


The day we learned about trigo graphs & cast diagrams I was ill. :frown:

The book sucks. Know of anywhere that I can learn this?
Original post by TeeEm
too late
Must you keep posting unneeded and often crudely sarcastic comments on every thread? This really is unacceptable.
Reply 7
Original post by Naruke
The day we learned about trigo graphs & cast diagrams I was ill. :frown:

The book sucks. Know of anywhere that I can learn this?


Here you go.
Reply 8
Original post by constellarknight
Must you keep posting unneeded and often crudely sarcastic comments on every thread? This really is unacceptable.


I am sorry you feel so...
I beginning to dislike you ...
Original post by TeeEm
I am sorry you feel so...
I beginning to dislike you ...


The honour... By the way, your IYGB Special Papers are riddled with errors. For example, Paper Z Question 2. The fact that as written, the denominator is sqrt(x+sqrt(x^2-1)) makes it impossible. What was intended was presumably sqrt(x)+sqrt(x^2-1) (note that the two square roots are separate, rather than the second being within the first), which is indeed what you start out with in your solution. I could go on, but I'd rather not. Anyway mate, the point is whatever ego and perceived reputation you may have built up for yourself, you're a nobody. Have a good day!
Reply 10
Original post by constellarknight
The honour... By the way, your IYGB Special Papers are riddled with errors. For example, Paper Z Question 2. The fact that as written, the denominator is sqrt(x+sqrt(x^2-1)) makes it impossible. What was intended was presumably sqrt(x)+sqrt(x^2-1) (note that the two square roots are separate, rather than the second being within the first), which is indeed what you start out with in your solution. I could go on, but I'd rather not. Anyway mate, the point is whatever ego and perceived reputation you may have built up for yourself, you're a nobody. Have a good day!


you clearly have issues
Reply 11
Original post by Zacken
Here you go.


Thanks!

Another question: Eliminate θ\theta to give an equation relating x and y:

c) x=Sinθ x = Sin \theta , y=Cos2θy = Cos^2 \theta

Would this be ok?

Sinθ=1Cos2θSin \theta = \sqrt1-Cos^2\theta

x=1y x = \sqrt1-y

I don't know how to extend the square root in latex but you catch my drift
Reply 12
Original post by Naruke
Thanks!

Another question: Eliminate θ\theta to give an equation relating x and y:

c) x=Sinθ x = Sin \theta , y=Cos2θy = Cos^2 \theta

Would this be ok?

Sinθ=1Cos2θSin \theta = \sqrt1-Cos^2\theta

x=1y x = \sqrt1-y

I don't know how to extend the square root in latex but you catch my drift


1. If you want to extend the square root use \sqrt{1-y}, i.e: put the curly braces around it.

2. That would be okay, I suppose...

3. It's a very ugly answer, x2+y=1x^2 + y = 1 is the same thing and looks much nicer, but I doubt you'd lose marks for writing down what you have. So yes, it's perfectly correct.
Reply 13
Original post by Zacken
1. If you want to extend the square root use \sqrt{1-y}, i.e: put the curly braces around it.

2. That would be okay, I suppose...

3. It's a very ugly answer, x2+y=1x^2 + y = 1 is the same thing and looks much nicer, but I doubt you'd lose marks for writing down what you have. So yes, it's perfectly correct.


Thank you sir
Original post by TeeEm
you clearly have issues


Thanks you narcissistic piece of ****.
Reply 15
Original post by Naruke
Thank you sir


Oi, I'm an A-Level student just like yourself! :biggrin:
Reply 16
Original post by constellarknight
Thanks you narcissistic piece of ****.


Did you have a bad day ?
I remember you lost it again in another thread the other day ...
Different username but identical persona...
How old are you please, if you do not mind me asking?
Reply 17
Original post by Zacken
Oi, I'm an A-Level student just like yourself! :biggrin:


If I consider you an A level student, it will just lower my self - esteem.

I mean, I've been trying to grow a beard for 6 months but you make it seem light work.

:frown:
Reply 18
Original post by Naruke
If I consider you an A level student, it will just lower my self - esteem.


Nah, don't be silly. I'd wager your results will be better than mine, for real. :smile:

I mean, I've been trying to grow a beard for 6 months but you make it seem light work.

:frown:


:rofl: :rofl: Seriously though, it's great that you're coming on here and asking questions, shows that you want to understand everything and that's going to get you far. :yep:
Original post by TeeEm
Did you have a bad day ?
I remember you lost it again in another thread the other day ...
Different username but identical persona...
How old are you please, if you do not mind me asking?


I do mind you asking. Also what "different username?" I demand that you desist from these libellous accusations at once.
me postulando maxime vexas, etiam quid dicere conaris verbis "nomen alium"? posco igitur ut statim desistas affirmare famosa haec.
(edited 8 years ago)

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