# Maths Ext 2 Predictions/Thoughts (1 Viewer)

#### s97127

##### Member
Almost...except you forgot the binomial coefficients.

Also, generally speaking, proving something is positive to prove it is non-zero doesn’t always work (it works in this case though). You can have expressions that can be positive or negative but is non-zero.
Thanks. I forgot about coefficients. Regarding the other comment, it only works if all the terms are either greater/lesser than 0.

#### uart

##### Member
same omission of detail as above in the locus question
Also, neither the Itute nor TL solns provided a satisfactory explanation for why the negative $\bg_white \sqrt{5}$ solution was invalid in Q14c(ii).
$\bg_white \cos^2{(\pi/10)} = \sqrt{\frac{5 \pm \sqrt{5}}{8} }$

Itute gave no reasoning for selection of the +ive inner surd, while TL reasoned that cos^2() must be positive (but clearly the [$\bg_white 5 - \sqrt{5}$] solution is also positive).

My reasoning is that where we subst,
$\bg_white \cos(5 \theta) = \cos(\pi/2) = 0$,
in order to get the polynomial, we could equally well have substituted
$\bg_white \cos(5 \theta) = \cos(3\pi/2) = 0$,
to get the same equation.

So the polynomial is equally valid for solving for, $\bg_white x = \cos(3\pi/10)$, as well as some other non acute angles (which clearly can be ignored).

So the two acute solutions,
$\bg_white \cos^2{(\cdot)} = \sqrt{\frac{5 \pm \sqrt{5}}{8} }$
correspond to solutions for cos(pi/10) and cos(3pi/10), and since cos() is a decreasing function on (0,pi), then the larger value must correspond to cos(pi/10).

Did anyone else use this reasoning?

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#### notme123

##### Well-Known Member
Also, neither the Itute nor TL solns provided a satisfactory explanation for why the negative $\bg_white \sqrt{5}$ solution was invalid in Q14c(ii).
$\bg_white \cos^2{(\pi/10)} = \sqrt{\frac{5 \pm \sqrt{5}}{8} }$

Itute gave no reasoning for selection of the +ive inner surd, while TL reasoned that cos^2() must be positive (but clearly the [$\bg_white 5 - \sqrt{5}$] solution is also positive).

My reasoning is that where we subst,
$\bg_white \cos(5 \theta) = \cos(\pi/2) = 0$,
in order to get the polynomial, we could equally well have substituted
$\bg_white \cos(5 \theta) = \cos(3\pi/2) = 0$,
to get the same equation.

So the polynomial is equally valid for solving for, $\bg_white x = \cos(3\pi/10)$, as well as some other non acute angles (which clearly can be ignored).

So the two acute solutions,
$\bg_white \cos^2{(\cdot)} = \sqrt{\frac{5 \pm \sqrt{5}}{8} }$
correspond to solutions for cos(pi/10) and cos(3pi/10), and since cos() is a decreasing function on (0,pi), then the larger value must correspond to cos(pi/10).

Did anyone else use this reasoning?
i didnt justify it at all i think whoops

#### jyu

##### Member
Also, neither the Itute nor TL solns provided a satisfactory explanation for why the negative $\bg_white \sqrt{5}$ solution was invalid in Q14c(ii).
$\bg_white \cos^2{(\pi/10)} = \sqrt{\frac{5 \pm \sqrt{5}}{8} }$

Itute gave no reasoning for selection of the +ive inner surd, while TL reasoned that cos^2() must be positive (but clearly the [$\bg_white 5 - \sqrt{5}$] solution is also positive).

My reasoning is that where we subst,
$\bg_white \cos(5 \theta) = \cos(\pi/2) = 0$,
in order to get the polynomial, we could equally well have substituted
$\bg_white \cos(5 \theta) = \cos(3\pi/2) = 0$,
to get the same equation.

So the polynomial is equally valid for solving for, $\bg_white x = \cos(3\pi/10)$, as well as some other non acute angles (which clearly can be ignored).

So the two acute solutions,
$\bg_white \cos^2{(\cdot)} = \sqrt{\frac{5 \pm \sqrt{5}}{8} }$
correspond to solutions for cos(pi/10) and cos(3pi/10), and since cos() is a decreasing function on (0,pi), then the larger value must correspond to cos(pi/10).

Did anyone else use this reasoning?
itute stated cos@=cospi/10=0.951approx so cannot be sqrt((5-sqrt5)/8)

##### -insert title here-
itute stated cos@=cospi/10=0.951approx so cannot be sqrt((5-sqrt5)/8)
ah yes the engineer's method

#### jyu

##### Member
ah yes the engineer's method
Show that 117^(0.4)+pi^(5/3)-e^2.5 is positive.

#### Siwel

##### Active Member
ah yes the engineer's method
is evaluating it and saying both are the same number not enough to prove it?

#### Trebla

is evaluating it and saying both are the same number not enough to prove it?
It's a circular argument. Prove that this trig result holds. Oh look, they happen to equal on a calculator therefore it is proven!

#### Siwel

##### Active Member
It's a circular argument. Prove that this trig result holds. Oh look, they happen to equal on a calculator therefore it is proven!
just thought it was kinda like the roots thing, if this polynomial can be tan3theta solve for theta then find the roots, then by plugging in the numbers to see which ones are the distinct roots.

#### Trebla

just thought it was kinda like the roots thing, if this polynomial can be tan3theta solve for theta then find the roots, then by plugging in the numbers to see which ones are the distinct roots.
But plugging in numbers to see what the exact trig value is lacks proper rigour, as you have effectively found the answer by direct computation rather than by proof. It is the same as claiming you have proven the exact value of something simply because the calculator says they're equal rather than by logical arguments.

#### uart

##### Member
It looks dirty
But plugging in numbers to see what the exact trig value is lacks proper rigour,
Agreed that it looks dirty, but it's probably still valid.

The point is that it's not about proving that it is the correct answer, it's about proving that the other surd is not the correct answer.

It would of course be invalid to try to use a calculator to prove that it was equal, because whatever decimal precision you use, it can never prove equity to the surd. But in terms of simply ruling out the other solution (the only other feasible solution), then it's dirty but it's valid (in my opinion).

#### jyu

##### Member
It looks dirty

Agreed that it looks dirty, but it's probably still valid.

The point is that it's not about proving that it is the correct answer, it's about proving that the other surd is not the correct answer.

It would of course be invalid to try to use a calculator to prove that it was equal, because whatever decimal precision you use, it can never prove equity to the surd. But in terms of simply ruling out the other solution (the only other feasible solution), then it's dirty but it's valid (in my opinion).
pi/10 < pi/6, cos(pi/10) > sqrt(3)/2, easily shown that sqrt((5 - sqrt5)/8) < sqrt(3)/2 and sqrt((5 + sqrt5)/8) > sqrt(3)/2