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Can someone help me starting B. I know that you have to join the two equations in terms of X and y. But how can I use y+1/y. ? Thanks


Posted from TSR Mobile
Original post by Lilly1234567890
ImageUploadedByStudent Room1458401458.913003.jpg

Can someone help me starting B. I know that you have to join the two equations in terms of X and y. But how can I use y+1/y. ? Thanks


Posted from TSR Mobile


You are given that something is equal to 2sectheta and something else is equal to 2cosectheta... can you think of any (not immediate) relationships between sectheta and costheta?
Reply 2
Avatar for joostan
joostan
13
Original post by Lilly1234567890
ImageUploadedByStudent Room1458401458.913003.jpg

Can someone help me starting B. I know that you have to join the two equations in terms of X and y. But how can I use y+1/y. ? Thanks


Posted from TSR Mobile

A possibly simpler alternative to what Sean FM is suggesting might be to consider taking the reciprocal of each equation and using a common identity:

Spoiler

(edited 8 years ago)
Reply 3
Avatar for Zacken
Zacken
22
I'd go with Joostan's method, I'm just going to add a little point that might trip someone up:

12sec⁡θ=1x+1x=1x2+1x=xx2+1≠1x+11x=1x+x\displaystyle \frac{1}{2\sec \theta} = \frac{1}{x + \frac{1}{x}} = \frac{1}{\frac{x^2 + 1}{x}} = \frac{x}{x^2 + 1} \neq \frac{1}{x} + \frac{1}{\frac{1}{x}} = \frac{1}{x} + x
Original post by Zacken
I'd go with Joostan's method, I'm just going to add a little point that might trip someone up:

12sec⁡θ=1x+1x=1x2+1x=xx2+1≠1x+11x=1x+x\displaystyle \frac{1}{2\sec \theta} = \frac{1}{x + \frac{1}{x}} = \frac{1}{\frac{x^2 + 1}{x}} = \frac{x}{x^2 + 1} \neq \frac{1}{x} + \frac{1}{\frac{1}{x}} = \frac{1}{x} + x

Or
[br]4y2y3+1−2y3=x3+1−2x24x2[br][br]\dfrac{4y^2}{y^3 + 1 - 2y^3} = \dfrac{x^3 + 1 - 2x^2}{4x^2}[br]
Original post by Zacken
I'd go with Joostan's method, I'm just going to add a little point that might trip someone up:

12sec⁡θ=1x+1x=1x2+1x=xx2+1≠1x+11x=1x+x\displaystyle \frac{1}{2\sec \theta} = \frac{1}{x + \frac{1}{x}} = \frac{1}{\frac{x^2 + 1}{x}} = \frac{x}{x^2 + 1} \neq \frac{1}{x} + \frac{1}{\frac{1}{x}} = \frac{1}{x} + x


thankssssss.
thanks allll
got it
Original post by joostan
A possibly simpler alternative to what Sean FM is suggesting might be to consider taking the reciprocal of each equation and using a common identity:

Spoiler



That is exactly what I was suggesting :tongue: just in a way that gives the OP more room to think for themselves :tongue:
Reply 8
Avatar for joostan
joostan
13
Original post by SeanFM
That is exactly what I was suggesting :tongue: just in a way that gives the OP more room to think for themselves :tongue:


Ah, my bad, but surely this uses the immediate relation between cos⁡(θ)\cos(\theta) and sec⁡(θ)\sec(\theta), which you suggested not to use. . .

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