Is my answer wrong? I got (1/2)a^2(2π−sin(2Ø)−2cos(Ø))

[Kulla_9289]

I got [imath]a^2\left(\pi -2ϕ\cos ^2\left(2ϕ\right)\right)[/imath]. By the way, the real question is, how did you know that solving the small triangle using trigonometry would give us the y value? How could you see that?

 

[Dr.Peterson]

I got [imath]a^2\left(\pi -2ϕ\cos ^2\left(2ϕ\right)\right)[/imath]. By the way, the real question is, how did you know that solving the small triangle using trigonometry would give us the y value? How could you see that?

Good question.

I looked for what parts of the picture involve y together with the desired variables a and [imath]\phi[/imath]:

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That isn't immediately obvious; but it becomes clearer when we identify other segments with the same lengths a and y:

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Once you see the isosceles triangle with sides a and y, and angle [imath]\phi[/imath], you can draw in the perpendicular and you have a useful right triangle.

But please show the details of your work; I think there is still an error.

 

[Kulla_9289]

I think I made a typing error. I meant to type πa² - 2a²ϕcos(ϕ)².

 

[Dr.Peterson]

I think I made a typing error. I meant to type πa² - 2a²ϕcos(ϕ)².

That still isn't right; one number in it is wrong.

I'm not sure we ever pointed out that this was wrong:

This is my final answer: a²π-y²∅. I do not know how one would remove the y. Ignore the other answers I said earlier.

If you would show your work, we wouldn't have to waste so much time.

 

[Kulla_9289]

Initially I got πa²-1/2(y²)(2ϕ). Later on I got y = 2a cosϕ. On substituting, I get πa² - 1/2(2a cosϕ)²(2ϕ). Simplifying it gets me to πa² -4a²ϕcos(ϕ)². Is this right?

 

[blamocur]

Initially I got πa²-1/2(y²)(2ϕ). Later on I got y = 2a cosϕ. On substituting, I get πa² - 1/2(2a cosϕ)²(2ϕ). Simplifying it gets me to πa² -4a²ϕcos(ϕ)². Is this right?

Looks good to me, but instead of [imath]\cos (\phi)^2[/imath] (which could also be interpreted as [imath]\cos(\phi^2)[/imath]) I'd typeset it as [imath]\cos^2\phi[/imath]