Hey Guys,

I am not sure how would you go about doing question 14? Anyone got any ideas?

Attachment 31336

Printable View

- Jul 24th 2014, 04:28 AMrainbowstarElimination
Hey Guys,

*I am not sure how would you go about doing question 14? Anyone got any ideas?*

Attachment 31336 - Jul 24th 2014, 04:35 AMebainesRe: Elimination
Strangely worded, but I think they want you to express y as a function of x. For example in the first one if you use the identity $\displaystyle \sin \theta = \sqrt {1- \cos^2 \theta}$ you can get an expression for y(x) which has no $\displaystyle \theta$ term in it.

- Jul 24th 2014, 04:48 AMrainbowstarRe: Elimination
That seems like a possible way. However, the answer is:

x square/a square + y square/bsquare = 1

Could you solve these trig equations simultaneously possibly?How would that work? - Jul 24th 2014, 05:03 AMdennydenglerRe: Elimination
Trigonometric identities needed for these 2 questions:

(a) cos^2(theta) + sin^2(theta) = 1. With that in mind, cos(theta) = x/a and sin(theta) = y/b. That's how you have (x/a)^2+(y/b)^2 = 1

(a) tan^2(theta) + 1 = sec^2(theta). With that in mind, tan(theta) = x/a and sec(theta) = y/b. Then you have (x/a)^2 + 1 = (y/b)^2.

Good day. - Jul 24th 2014, 05:03 AMebainesRe: Elimination
Did you try the approach I suggested? It gets to the same answer:

$\displaystyle y = b \sin \theta$

$\displaystyle y = b \sqrt {1- \cos^2 \theta} $

$\displaystyle y^2 = b^2(1- \cos^2 \theta) $

$\displaystyle y^2 = b^2(1 - (\frac x a)^2) $

$\displaystyle \frac {y^2}{b^2} + \frac {x^2}{a^2} = 1 $ - Jul 24th 2014, 05:32 AMrainbowstarRe: Elimination
Thanks ebaines and dennydengler