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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: :) on January 14, 2009, 03:58:51 pm

Title: Circular Function Questions
Post by: :) on January 14, 2009, 03:58:51 pm: 1) How would you simplify this...

$(4sin^2x - 4sin^4x) /sin2x$

2) Let sinx = 0.6, x is within the domain [pi/2,pi] and tan y=2.4, y is within the domain [0,pi/2]

Find the value of tan(x + 2y)

Thanks :)
Title: Re: Circular Function Questions
Post by: Mao on January 14, 2009, 11:06:59 pm: $(1)=\frac{4\sin^2 x - 4\sin^4 x}{\sin 2x} = \frac{4\sin^2 x - 4\sin^4 x}{2\sin x\cdot \cos x}=\frac{2\sin x\cdot (1-\sin^2 x)}{\cos x} = 2\sin x \cdot \cos x=\sin 2x$
Title: Re: Circular Function Questions
Post by: kj_ on January 18, 2009, 11:01:22 pm: This seems a bit whack, so so correct me if I'm wrong:

From your compound angle formula, we recognize that $tan(x+2y)=\frac{tan(x)+tan(2y)}{1-tan(x).tan(2y)}$
Thus, we have to determine values for $tan(x)$ , and $tan(2y)$

We're given that $sin(x) = 0.6 = \frac{3}{5}$ , and that it's in the 2nd quadrant (as restricted by your domain).
Thus, from a diagram or from the identity $sin^2(x) + cos^2(x) = 1$ , we determine that
$tan(x) = -\frac{3}{4}$ (note the negative as it's in the 2nd quadrant)

We're also given that $tan(y)=2.4=\frac{12}{5}$
Thus, using your double angle formula, $tan(2y)=\frac{2tan(y)}{1-tan^2(y)}$
= $\frac {2.\frac{12}{5}}{1-\frac{144}{25}}$
= $\frac{24}{5}.-\frac{25}{119}$
$tan(2y)=-\frac {120}{119}$

And thus, substituting these two values into your compound angle formula, we obtain:
$tan(x+2y)=\frac{tan(x)+tan(2y)}{1-tan(x).tan(2y)}$
$=>tan(x+2y)=\frac{-\frac{3}{4}-\frac {120}{119}}{1-\frac{3}{4}.\frac {120}{119}}$
$= -\frac{837}{116}$