Mathematics Challenge Marathon

[RuiAce]

Here, I will occasionally post challenge questions for the daring 2U (and above) student to attempt.

Questions are not intended to reflect the scope of the difficulty in actual HSC questions, however may be completed using only knowledge taught in the course. Spoilers are intended to reveal what topics to draw knowledge from when a genuine, unaided attempt has been unsuccessful.

I invite everyone to also post their own questions at their discretion.

Spoiler

Required knowledge: Preliminary Basic Arithmetic & Algebra, Preliminary The Quadratic Function

[brenden]

Here, I will occasionally post challenge questions for the daring 2U (and above) student to attempt.

Questions are not intended to reflect the scope of the difficulty in actual HSC questions, however may be completed using only knowledge taught in the course. Spoilers are intended to reveal what topics to draw knowledge from when a genuine, unaided attempt has been unsuccessful.

I invite everyone to also post their own questions at their discretion.

Spoiler

Required knowledge: Preliminary Basic Arithmetic & Algebra, Preliminary The Quadratic Function

Awesome!!

[Happy Physics Land]

Good stuff Rui, you can finally share your wisdom with this awesome community!

[KoA]

Solution

[keltingmeith]

Solution

Your solution for the first isn't too bad! But, you've only proven that:



You need to have another shot at proving the other way round, methinks. (albeit, the other way round is a LOT more difficult, and requires more logical reasoning than straight up mathematics)

I love just how smoothly you put that into the second half of the question, though! Absolutely smashed it.

[RuiAce]

Solution

The second part was done well.

However, whilst the first part was done correctly, it was not done in the most elegant manner:



NEXT QUESTION


You may answer for 0°≤x≤360° if you have not been exposed to the radian measure of an angle.

Spoiler

Required knowledge: Preliminary Trigonometric Ratios

[Happy Physics Land]

memes

[jakesilove]

memes

(Image removed from quote.)

Gonna have to jump in there: Whilst it all looks like good maths, you divide by (sinx-cosx) between lines 2 and three. Whenever you divide by a term like that, you must explicitly state that (sinx-cosx) does not equal zero, and then using that equation figure out values for x that cannot exist (Maybe not in 2U, but since I know you do 4U...). This will likely change some of your final answers.

Jake

[RuiAce]

Gonna have to jump in there: Whilst it all looks like good maths, you divide by (sinx-cosx) between lines 2 and three. Whenever you divide by a term like that, you must explicitly state that (sinx-cosx) does not equal zero, and then using that equation figure out values for x that cannot exist (Maybe not in 2U, but since I know you do 4U...). This will likely change some of your final answers.

Jake

Yeah. The case that (sin(x)-cos(x))=0 has been disregarded in this case, and thus tan(x)=1 was ignored.

[Happy Physics Land]

Gonna have to jump in there: Whilst it all looks like good maths, you divide by (sinx-cosx) between lines 2 and three. Whenever you divide by a term like that, you must explicitly state that (sinx-cosx) does not equal zero, and then using that equation figure out values for x that cannot exist (Maybe not in 2U, but since I know you do 4U...). This will likely change some of your final answers.

Jake

Ahhhhhh ok thank you so much Jake for pointing that out!!! I'm actually quite unaware of this detail and I guess its gonna be fatal in an exam (better jot this down).

ok so if l consider sinx - cosx cannot equal 0
then sinx cannot equal cos x
then tan x cannot equal 1
then x cannot equal pi/4 and that shouldnt really matter
but since we are dividing by cos x on both sides
cos x cannot equal 0
hence x cannot equal to pi/2 or 3pi/2

So the final answer for x would be x = 0, pi and 2pi?

[RuiAce]

Ahhhhhh ok thank you so much Jake for pointing that out!!! I'm actually quite unaware of this detail and I guess its gonna be fatal in an exam (better jot this down).

ok so if l consider sinx - cosx cannot equal 0
then sinx cannot equal cos x
then tan x cannot equal 1
then x cannot equal pi/4 and that shouldnt really matter
but since we are dividing by cos x on both sides
cos x cannot equal 0
hence x cannot equal to pi/2 or 3pi/2

So the final answer for x would be x = 0, pi and 2pi?

The only thing that Jake justified was that this is true if and only if sin(x)-cos(x)≠0.

In this case, there is no exception to this.

[RuiAce]

NEXT QUESTION

Spoiler

Required knowledge: Preliminary Trigonometric Ratios, HSC Series

[nerdgasm]

Here's my attempt at a solution (rather handwavy, but I'm lazy):

I'll first assume that we're summing over k (seeing as I can't see an 'n' in either summation expression)
Note that the restriction on the value of theta implies that both sin(theta) and cos(theta) are positive. From there, it is easy to see that their powers must be as well; therefore both S and C are non-zero.

We may recognise S and C as infinite geometric series with ratio term sin^2(theta) and cos^2(theta) respectively. Note also that the value of sin(theta) and cos(theta) is between 0 and 1, and hence the value of sin^2(theta) and cos^2(theta) is also between 0 and 1. Therefore, as our infinite geometric series has ratio term between 0 and 1, we may use the formula for sum of an infinite geometric series.

If we do this, we should end up with S = 1/(1-sin^2(theta)) and C = 1/(1-cos^2(theta)). The denominators are non-zero again by virtue of sin^2(theta) and cos^2(theta) lying strictly between 0 and 1.

Therefore, 1/S + 1/C = 2 - sin^2(theta) - cos^2(theta) = 1, as required.

[RuiAce]

Here's my attempt at a solution (rather handwavy, but I'm lazy):

I'll first assume that we're summing over k (seeing as I can't see an 'n' in either summation expression)
Note that the restriction on the value of theta implies that both sin(theta) and cos(theta) are positive. From there, it is easy to see that their powers must be as well; therefore both S and C are non-zero.

We may recognise S and C as infinite geometric series with ratio term sin^2(theta) and cos^2(theta) respectively. Note also that the value of sin(theta) and cos(theta) is between 0 and 1, and hence the value of sin^2(theta) and cos^2(theta) is also between 0 and 1. Therefore, as our infinite geometric series has ratio term between 0 and 1, we may use the formula for sum of an infinite geometric series.

If we do this, we should end up with S = 1/(1-sin^2(theta)) and C = 1/(1-cos^2(theta)). The denominators are non-zero again by virtue of sin^2(theta) and cos^2(theta) lying strictly between 0 and 1.

Therefore, 1/S + 1/C = 2 - sin^2(theta) - cos^2(theta) = 1, as required.

My apologies for the misleading 'n' term. But yes, correct!

It may have however, been neater to convert 1-sin²θ to cos²θ before applying the reciprocal. However this doesn't damage the elegance of the process.
____________________________________

NEXT QUESTION:

I didn't like my original question so I have replaced it:

Spoiler

Required knowledge: Preliminary Basic Arithmetic and Algebra, HSC Series

[juliakate_99]

NEXT QUESTION


You may answer for 0°≤x≤360° if you have not been exposed to the radian measure of an angle.

Spoiler

Required knowledge: Preliminary Trigonometric Ratios

Would the answer be 2pi/3 and 4pi/3?
( I couldn't upload a photo of my working, but basically divided both sides of the equation by sinx -cosx giving me sin^2x-cos^2x=1. Rearranged to get cos^2x-sin^2x=-1, therefore cos2x=-1, cos x=-1/2.