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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Milkshake on September 11, 2010, 03:22:47 pm

Title: Circular functions
Post by: Milkshake on September 11, 2010, 03:22:47 pm: tan(x) = a in the second quadrant.
what is sin (x)?
Title: Re: Circular functions
Post by: 98.40_for_sure on September 11, 2010, 03:28:12 pm: $sin(x)=-\frac{a}{\sqrt{1+a^{2}}}$
Is that right?
Title: Re: Circular functions
Post by: Milkshake on September 11, 2010, 03:30:20 pm: think so, the book gives a different answer though. can you show your working out please?
Title: Re: Circular functions
Post by: 98.40_for_sure on September 11, 2010, 03:32:17 pm: Construct a right angled triangle with sides -a, 1, $\sqrt{1+a^{2}}$
My reasoning for it being -a and not +a is because tan is positive in second quadrant?
What does the book say?
Title: Re: Circular functions
Post by: Milkshake on September 11, 2010, 03:35:33 pm: the book says the answer you got, except it was positive, so a/(1+a^2)^0.5
Title: Re: Circular functions
Post by: 98.40_for_sure on September 11, 2010, 03:43:23 pm: Hmm i'm not sure, i mean you could construct a triangle with -1 and a. Then you would get the book's answer. but... you can't have a length of -1 can you?
Wait for truetears to come on and solve this :P
Title: Re: Circular functions
Post by: 98.40_for_sure on September 11, 2010, 03:48:00 pm: 2nd attempt:

let $a=-x$

Construct triangle of sides -x, 1, $\sqrt{1+x^{2}}$

$a^{2}=x^{2}$

$sin(x)=\frac{-x}{\sqrt{1+x^{2}}}$

$=\frac{a}{\sqrt{1+a^{2}}}$
Title: Re: Circular functions
Post by: TrueTears on September 11, 2010, 05:41:15 pm: Quote from: Milkshake on September 11, 2010, 03:22:47 pm
tan(x) = a in the second quadrant.
what is sin (x)?

As requested.

$\frac{\sin(x)}{\cos(x)} = a$

$\frac{\sin^2(x)}{\cos^2(x)}=a^2$

$\frac{\sin^2(x)}{1-\sin^2(x)}=a^2$

(Putting in an extra step as requested: let $y = \sin(x) \implies \frac{y^2}{1-y^2} = a^2$ now solve for $y$ )

$\sin(x) = \frac{a}{\sqrt{a^2+1}}$ as $\sin(x) \ge 0$ in second quadrant.

I assume this is how the solutions did it, they probably forgot a itself was negative lol

Actually this question is quite stupid...

Um $\tan(x) \le 0$ in second quadrant so $a \le 0$

Thus if we have $\sin(x) = \frac{a}{\sqrt{a^2+1}}$ and $a \le 0$ then we should have $\sin(x) = -\frac{a}{\sqrt{a^2+1}}$ since $\sqrt{a^2+1} > 0 \ \forall \ a \in \mathbb{R}$
Title: Re: Circular functions
Post by: Martoman on September 11, 2010, 06:49:19 pm: Quote from: 99.95_for_sure on September 11, 2010, 03:28:12 pm
$sin(x)=-\frac{a}{\sqrt{1+a^{2}}}$
Is that right?

This is right. Its what i did. :smitten:
Title: Re: Circular functions
Post by: Furbob on September 15, 2010, 08:29:54 pm: can I throw in a question since its somewhat relevant

Q : find cos(3x) in terms of cos(x); hint: cos(3x) = cos(x+2x)

I was thinking about whether to use the addition or double angle formula but hmm....
Title: Re: Circular functions
Post by: 98.40_for_sure on September 15, 2010, 08:40:13 pm: Is the answer $4cos^{3}(x)-3cos(x)$ ?
i'm not sure what it means by in terms of cos(x)

EDIT: working below

$\cos (3x)=\cos(2x+x)$
$=\cos(2x)\cos(x)-\sin(2x)\sin(x)$
$=\left (2\cos^{2}(x)-1 \right )\cos(x)-2\sin^{2}(x)\cos(x)$
$=2\cos^{3}(x)-\cos(x)-2\cos(x)+2\cos^{3}(x)$
$=4\cos^{3}(x)-3\cos(x)$
Title: Re: Circular functions
Post by: Mao on September 15, 2010, 08:44:36 pm: Use the addition formula. Double angle formula can only be used when it's '2x' in there.