Author Topic: 4U Maths Question Thread (Read 665237 times) Tweet Share

Shadowxo · « **Reply #900 on:** February 28, 2017, 12:16:19 am »

Quote from: cutiepie30 on February 27, 2017, 11:51:30 pm

Thanks alot RuiAce !!

i get how you get equation 1 and 2

but i dont get how you added equations 1 and 2 and simplified from there , could you please do those steps in more depth

Also for your last case is it meant to be pi/4 for all real number and not pi/2?

$\int_0^{\frac\pi{2}}\frac{(\cos x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx + \int_0^{\frac\pi{2}}\frac{(\sin x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx \\ = \int_0^{\frac\pi{2}}\frac{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx$
Same denominator and boundaries , so it's just adding two fractions. The top is then equal to the bottom so it cancels out and equals 1.

cutiepie30 · « **Reply #901 on:** February 28, 2017, 12:33:42 am »

Quote from: Shadowxo on February 28, 2017, 12:16:19 am

$\int_0^{\frac\pi{2}}\frac{(\cos x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx + \int_0^{\frac\pi{2}}\frac{(\sin x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx \\ = \int_0^{\frac\pi{2}}\frac{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx$
Same denominator and boundaries , so it's just adding two fractions. The top is then equal to the bottom so it cancels out and equals 1.

Thanks Shadowxo

just wondering in the end it should be pi/4 for all cases in reference to rui ace's orignal solution ?

Shadowxo · « **Reply #902 on:** February 28, 2017, 12:57:19 am »

Quote from: cutiepie30 on February 28, 2017, 12:33:42 am

Thanks Shadowxo

just wondering in the end it should be pi/4 for all cases in reference to rui ace's orignal solution ?

I believe so

michaelalt · « **Reply #903 on:** February 28, 2017, 08:58:24 am »

Can someone help me with this? I don't know how to make it true for n=1

RuiAce · « **Reply #904 on:** February 28, 2017, 12:49:08 pm »

Quote from: michaelalt on February 28, 2017, 08:58:24 am

Can someone help me with this? I don't know how to make it true for n=1

$\text{Combine de Moivre's theorem with the difference of two squares factorisation}\\ \begin{align*}RHS&=\frac{1-\text{cis } 2\theta}{1-\text{cis } \theta}\\ &= \frac{1-(\text{cis } \theta)^2}{1-\text{cis } \theta}\\ &= \frac{(1-\text{cis } \theta)(1+\text{cis } \theta)}{1-\text{cis } \theta}\\ &= 1+\text{cis } \theta\\ &= LHS\end{align*}$

cutiepie30 · « **Reply #905 on:** February 28, 2017, 06:48:42 pm »

Quote from: RuiAce on February 27, 2017, 10:23:21 pm

$\text{Assuming that was meant to be }(\tan x)^{\sqrt \pi}$
$\text{Let }I=\int_{0}^{\frac{\pi }{2}}\frac{1}{1+(\tan x)^{\sqrt{\pi }}} dx\\ \text{Since }\tan x=\frac{\sin x}{\cos x}\text{, multiplying top}\\ \text{and bottom by }(\cos x)^{\sqrt \pi}\text{ gives}\\ I=\int_0^{\frac\pi{2}}\frac{(\cos x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx$
$\text{Call that equation 1}\\ \text{In equation 1, perform the trick }\int_0^a f(x)dx=\int_0^a f(a-x)dx\\ \text{with }a=\frac{\pi}{2}\\ \text{Then, noting the complementary identities e.g. }\sin \left(\frac{\pi}2-x\right)=\cos x\\ \text{We have }I=\int_0^{\frac\pi{2}}\frac{(\sin x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx\\ \text{Call that equation 2}$
$\text{Adding equations 2 and 1 tidies everything up}\\ \begin{align*}2I&=\int_0^{\frac\pi{2}}\frac{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}{(\sin x)^{\sqrt\pi}+(\cos x)^{\sqrt\pi}}dx\\ I&=\int_0^{\frac\pi 2}dx\\ &= \frac{\pi}{4}\end{align*}$
_____
$\text{In fact, it can be shown that for all real numbers }\alpha\\ \int_0^{\frac\pi 2}\frac{1}{1+\tan^\alpha x}dx=\frac{\pi}2$

Hey ruiAce , just a question in regards to your solution?

How did you know and get this trick ? How did you figure it out / get to it? $\text{In equation 1, perform the trick }\int_0^a f(x)dx=\int_0^a f(a-x)dx\\ \text{with }a=\frac{\pi}{2}\\$

Thanks

!

RuiAce · « **Reply #906 on:** February 28, 2017, 07:03:34 pm »

Quote from: cutiepie30 on February 28, 2017, 06:48:42 pm

Hey ruiAce , just a question in regards to your solution?

How did you know and get this trick ? How did you figure it out / get to it? $\text{In equation 1, perform the trick }\int_0^a f(x)dx=\int_0^a f(a-x)dx\\ \text{with }a=\frac{\pi}{2}\\$

Thanks !

$\text{This trick is standard for 4U and the proof needs to be memorised}\\ \text{In the RHS, perform the substitution }u=a-x$
$\begin{align*}\int_0^a f(a-x)dx &= \int_a^0 f(u)(-du)\\ &= \int_0^a f(u)du\\ &= \int_0^a f(x)dx\end{align*}$
$\text{The substitution flips the boundaries but the negative sign flips it back}\\ \text{Then, the rule that }\int_a^bf(x)dx = \int_a^bf(u)du\text{ should make intuitive sense}\\ \text{When performing definite integration, the variables vanish, leaving the same answer.}$

michaelalt · « **Reply #907 on:** February 28, 2017, 07:32:54 pm »

Quote from: RuiAce on February 28, 2017, 12:49:08 pm

$\text{Combine de Moivre's theorem with the difference of two squares factorisation}\\ \begin{align*}RHS&=\frac{1-\text{cis } 2\theta}{1-\text{cis } \theta}\\ &= \frac{1-(\text{cis } \theta)^2}{1-\text{cis } \theta}\\ &= \frac{(1-\text{cis } \theta)(1+\text{cis } \theta)}{1-\text{cis } \theta}\\ &= 1+\text{cis } \theta\\ &= LHS\end{align*}$

Thanks Rui! You make it so simple haha.
I get really confused on the n=k+1 part, any help would be greatly appreciated!

RuiAce · « **Reply #908 on:** February 28, 2017, 07:44:31 pm »

Quote from: michaelalt on February 28, 2017, 07:32:54 pm

Thanks Rui! You make it so simple haha.
I get really confused on the n=k+1 part, any help would be greatly appreciated!

$\begin{align*}&\quad1+\text{cis }\theta+\text{cis }2\theta+\dots+\text{cis }k\theta+\text{cis }[(k+1)\theta]\\&= \frac{1-\text{cis }[(k+1)\theta]}{1-\text{cis }\theta}+\text{cis }[(k+1)\theta] \tag{assumption}\\ &= \frac{1-\text{cis }[(k+1)\theta]+\left(\text{cis }[(k+1)\theta]-\text{ cis}\theta \text{ cis }[(k+1)\theta ]\right)}{1-\text{cis }\theta}\tag{combining into one fraction}\\ &= \frac{1-\text{cis }\theta \text{ cis }[(k+1)\theta]}{1-\text{cis }\theta}\\ &= \frac{1-\text{cis }[(k+2)\theta]}{1-\text{cis }\theta}\tag{*}\end{align*}$
$(*) -\text{Recall that }\text{cis }\alpha \times \text{cis }\beta = \text{cis }(\alpha+\beta)$

michaelalt · « **Reply #909 on:** February 28, 2017, 08:00:50 pm »

Quote from: RuiAce on February 28, 2017, 07:44:31 pm

$\begin{align*}&\quad1+\text{cis }\theta+\text{cis }2\theta+\dots+\text{cis }k\theta+\text{cis }[(k+1)\theta]\\&= \frac{1-\text{cis }[(k+1)\theta]}{1-\text{cis }\theta}+\text{cis }[(k+1)\theta] \tag{assumption}\\ &= \frac{1-\text{cis }[(k+1)\theta]+\left(\text{cis }[(k+1)\theta]-\text{ cis}\theta \text{ cis }[(k+1)\theta ]\right)}{1-\text{cis }\theta}\tag{combining into one fraction}\\ &= \frac{1-\text{cis }\theta \text{ cis }[(k+1)\theta]}{1-\text{cis }\theta}\\ &= \frac{1-\text{cis }[(k+2)\theta]}{1-\text{cis }\theta}\tag{*}\end{align*}$
$(*) -\text{Recall that }\text{cis }\alpha \times \text{cis }\beta = \text{cis }(\alpha+\beta)$

You are so helpful. Thank you so much for taking time to write it all out Rui!!
If you have time, would possibly be able to give me some help on this question as well? Thank you for all your help so far!!!

RuiAce · « **Reply #910 on:** February 28, 2017, 08:11:48 pm »

Quote from: michaelalt on February 28, 2017, 08:00:50 pm

You are so helpful. Thank you so much for taking time to write it all out Rui!!
If you have time, would possibly be able to give me some help on this question as well? Thank you for all your help so far!!!

Hint: Part c) is the useful part in the worked example. Commence by $\text{Let }x=\tan \theta$ then play around with the algebra until you can use the triple-angle identity like they did

Feel free to post any progress if you get a bit lost halfway through

michaelalt · « **Reply #911 on:** February 28, 2017, 08:30:06 pm »

Quote from: RuiAce on February 28, 2017, 08:11:48 pm

Hint: Part c) is the useful part in the worked example. Commence by $\text{Let }x=\tan \theta$ then play around with the algebra until you can use the triple-angle identity like they did

Feel free to post any progress if you get a bit lost halfway through

OK... I've gotten abit lost. HAHA.

kiwiberry · « **Reply #912 on:** February 28, 2017, 08:37:18 pm »

Quote from: michaelalt on February 28, 2017, 08:30:06 pm

OK... I've gotten abit lost. HAHA.

Nearly there!

From your last line:
$\frac{3\tan \theta - \tan^3 \theta}{\sqrt3(1-3\tan^2 \theta)} = 1 \\ \frac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta} = \sqrt3 \\ \tan 3\theta = \sqrt3$

michaelalt · « **Reply #913 on:** February 28, 2017, 08:43:31 pm »

Quote from: kiwiberry on February 28, 2017, 08:37:18 pm

Nearly there! From your last line:
$\frac{3\tan \theta - \tan^3 \theta}{\sqrt3(1-3\tan^2 \theta)} = 1 \\ \frac{3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta} = \sqrt3 \\ \tan 3\theta = \sqrt3$

ahh ! I get it haha, thanks heaps!

michaelalt · « **Reply #914 on:** February 28, 2017, 09:05:44 pm »

I have no idea how to do part b. and c. of this question. I've tried watching videos but the examples are all different, and theres no pattern to them so I'm having a hard time working them out...

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