Author Topic: 3U Maths Question Thread (Read 1518306 times) Tweet Share

bsdfjnlkasn · « **Reply #1590 on:** March 05, 2017, 01:23:32 pm »

Hey I have a few questions about the Further Trigonometry topic because I really just don't understand it so if anyone could help me out I'd really appreciate it

Q13e

Q15c:

Q4d:

Q7b:

Shadowxo · « **Reply #1591 on:** March 05, 2017, 04:10:25 pm »

Quote from: bsdfjn;lkasn on March 05, 2017, 01:23:32 pm

Hey I have a few questions about the Further Trigonometry topic because I really just don't understand it so if anyone could help me out I'd really appreciate it

Q13e
(Image removed from quote.)

Q15c:

(Image removed from quote.)

Q4d:

(Image removed from quote.)

Q7b:

(Image removed from quote.)

I'll just do the first one

13e)
$\frac{ \sin 2\theta - \cos 2\theta +1}{ \sin 2\theta + \cos 2\theta -1} \\ =\frac{2\sin \theta \cos \theta -(1-2 \sin^2 \theta) +1}{2 \sin \theta \cos \theta + (1-2 \sin^2 \theta)-1} \\ =\frac{2\sin\theta \cos\theta+2sin^2\theta}{2\sin\theta\cos\theta-2sin^2\theta} \\ =\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\\=\frac{1+\tan\theta}{1-\tan\theta} \\=\frac{\tan{\frac{\pi}{4}}+\tan\theta}{1-\tan{\frac{\pi}{4}}\tan\theta}\\=\tan({\theta+\frac{\pi}{4}})$
For questions like these, keep in mind what you want to prove - in this case, it's tan(a+b), so keep in mind the form you need it to be in. Try to simplify it as much as you can, and it's often good to get it into the same form - eg a single theta in this case, and not 2theta, and try to get rid of constants, or get rid of them via an identity.
15c)
$\frac{x}{2}=\tan{\frac{\theta}{2}} \\ y=\cos\theta\\y=2cos^2\frac{\theta}{2}-1 \\ \frac{y+1}{2}=cos^2\frac{\theta}{2} \\ \sec^2\frac{\theta}{2}=\frac{2}{y+1}\\1+ \tan^2\theta=\sec^2\theta \\ 1+\frac{x^2}{4}=\frac{2}{y+1}\\\frac{2}{y+1}-\frac{x^2}{4}=1$
For this, just make them in the same form (theta/2 in this case, instead of theta) and see the relevant trig identity to put them together
Hope this helps a bit

EDIT: Added 15c) answer. Also I'm a bit rusty so sorry for any errors

Syndicate · « **Reply #1592 on:** March 05, 2017, 04:30:56 pm »

Quote from: bsdfjn;lkasn on March 05, 2017, 01:23:32 pm

Hey I have a few questions about the Further Trigonometry topic because I really just don't understand it so if anyone could help me out I'd really appreciate it

Q13e
(Image removed from quote.)

Q15c:

(Image removed from quote.)

Q4d:

(Image removed from quote.)

Q7b:

(Image removed from quote.)

Q7b:

Spoiler

$tan(2 \times \frac{7\pi}{8} ) = tan( \frac{7\pi}{4}) = -1$
$\implies \frac{2 \times 2tan(\frac{7 \pi}{8})}{1-tan^2(\frac{7\pi}{8})}=-1$
$\text{Let} \space tan(\frac{7 \pi}{8}) \space \text{= a}$
$\frac{2a}{1-a^2}=-1$
$2a=a^2-1$
$a^2-2a-1=0$
$a = \frac{2 \pm \sqrt{4+4}}{2}$
$a =1 \pm \sqrt{2}$
$\therefore tan(\frac{7 \pi}{8}) = 1 \pm \sqrt{2}$
$\text{since} \space tan(\frac{7 \pi}{8}) \space \text{is in the second quadrant}$
$tan(\frac{7 \pi}{8}) = 1- \sqrt{2}$

I hope this working out is valid in HSC 3u maths

RuiAce · « **Reply #1593 on:** March 05, 2017, 09:47:00 pm »

Quote from: Syndicate on March 05, 2017, 04:30:56 pm

Q7b:
Spoiler
$tan(2 \times \frac{7\pi}{8} ) = tan( \frac{7\pi}{4}) = -1$
$\implies \frac{2 \times 2tan(\frac{7 \pi}{8})}{1-tan^2(\frac{7\pi}{8})}=-1$
$\text{Let} \space tan(\frac{7 \pi}{8}) \space \text{= a}$
$\frac{2a}{1-a^2}=-1$
$2a=a^2-1$
$a^2-2a-1=0$
$a = \frac{2 \pm \sqrt{4+4}}{2}$
$a =1 \pm \sqrt{2}$
$\therefore tan(\frac{7 \pi}{8}) = 1 \pm \sqrt{2}$
$\text{since} \space tan(\frac{7 \pi}{8}) \space \text{is in the second quadrant}$
$tan(\frac{7 \pi}{8}) = 1- \sqrt{2}$

I hope this working out is valid in HSC 3u maths

This definitely is valid

Mahan · « **Reply #1594 on:** March 05, 2017, 11:23:02 pm »

If a,b,c > 0 and a+b = c. Then find the minimum value of the expression
$A = \frac{a^{4} + b^{4} + c^{4}}{(ab)^{2}}$

bsdfjnlkasn · « **Reply #1595 on:** March 06, 2017, 05:32:20 pm »

Hey there, I was wondering if someone could help me solve this

If cos(alpha)=2/3 what is cos4(theta)

Thank you!!

Mahan · « **Reply #1596 on:** March 06, 2017, 05:39:22 pm »

Quote from: bsdfjn;lkasn on March 06, 2017, 05:32:20 pm

Hey there, I was wondering if someone could help me solve this

If cos(alpha)=2/3 what is cos4(theta)

Thank you!!

You need to use double angle formula:
$cos(2\alpha) = 2cos^{2}(\alpha) -1$
then
$cos(2\alpha) = 2cos^{2}(\alpha) -1= 2(\frac{2}{3})^2 -1 = -\frac{1}{9}$
similarly
$cos(4\alpha) = 2cos^2(2\alpha) -1 = \frac{2}{81}-1 = -\frac{79}{81}$
I hope that helped!

bsdfjnlkasn · « **Reply #1597 on:** March 06, 2017, 05:47:38 pm »

Hey me again,

Could I please get some help with question 5 e?
Thank you

Mahan · « **Reply #1598 on:** March 06, 2017, 06:06:16 pm »

Quote from: bsdfjn;lkasn on March 06, 2017, 05:47:38 pm

Hey me again,

Could I please get some help with question 5 e?
Thank you

For this one you should use $sin(x-y) = sin(x) \ cos(y) - sin(y) \ cos(x)$ and the definition of tan in terms of sine and cosine:
$LHS = \frac{ \frac{sin(\theta) sin(\frac{\theta}{2} ) }{cos(\theta) cos(\frac{\theta}{2} } }{\frac{sin\theta}{cos\theta} - \frac{sin(\frac{\theta}{2}) }{cos(\frac{\theta}{2})}}= \frac{\frac{sin(\theta) sin(\frac{\theta}{2})}{cos(\theta) cos(\frac{\theta}{2})}}{\frac{sin(\theta) cos(\frac{\theta}{2}) -sin(\frac{\theta}{2}) cos(\theta)}{cos(\theta) cos(\frac{\theta}{2})}}$

$=\frac{sin\theta sin\frac{\theta}{2}}{sin\theta cos\frac{\theta}{2} - sin \frac{\theta}{2} cos\theta} = \frac{sin\theta sin\frac{\theta}{2}}{sin(\theta - \frac{\theta}{2}) }$ using the identity.
$= \frac{sin\theta sin\frac{\theta}{2}}{sin\frac{\theta}{2}} = sin\theta$

bsdfjnlkasn · « **Reply #1599 on:** March 06, 2017, 06:08:59 pm »

Quote from: Mahan on March 06, 2017, 06:06:16 pm

For this one you should use $sin(x-y) = sin(x) \ cos(y) - sin(y) \ cos(x)$ and the definition of tan in terms of sine and cosine:

$LHS = \frac{tan(\theta) tan(\frac{\theta}{2})}{tan(\theta) - tan(\frac{\theta}{2})} = \frac{\frac{sin(\theta) sin(\frac{\theta}{2}) }{cos(\theta) cos(\frac{\theta}{2})}{\frac{sin(\theta)}{ cos(\theta)} - \frac{sin(\frac{\theta}{2}) }{cos(\frac{\theta}{2}) }}$

$= \frac{\frac{sin(\theta) sin(\frac{\theta}{2})}{cos(\theta) cos(\frac{\theta}{2})}}{\frac{sin(\theta) cos(\frac{\theta}{2}) -sin(\frac{\theta}{2}) cos(\theta)}{cos(\theta) cos(\frac{\theta}{2})}}$

$=\frac{sin\theta sin\frac{\theta}{2}}{sin\theta cos\frac{\theta}{2} - sin \frac{\theta}{2} cos\theta} = \frac{sin\theta sin\frac{\theta}{2}}{sin(\theta - \frac{\theta}{2}) }$ using the identity.
$= \frac{sin\theta sin\frac{\theta}{2}}{sin\frac{\theta}{2}} = sin\theta$

Hey Mahan!! Thanks so much for tending to my questions so promptly they make a lot of sense

I was wondering if there had been a formatting malfunction in the LHS bar statement towards the top of the solution and if there was anyway to fix it - if not that's ok. Thanks again, i'll definitely have more questions soon

kiwiberry · « **Reply #1600 on:** March 06, 2017, 06:10:11 pm »

Quote from: bsdfjn;lkasn on March 06, 2017, 05:47:38 pm

Hey me again,

Could I please get some help with question 5 e?
Thank you

Another method - the question does say to use the t-formulae

We know that if $t=\tan \frac{\theta}{2}$, then
$\tan \theta = \frac{2t}{1-t^2}$
Subbing into the question,
$\begin{align*}\frac{\tan \theta \tan \frac{\theta}{2}}{\tan \theta -\tan \frac{\theta}{2}}&= \frac{\frac{2t}{1-t^2} \times t}{\frac{2t}{1-t^2}-t} \\&=\frac{2t^2}{2t-t(1-t^2)} \\ &=\frac{2t}{1+t^2} \\ &=\sin \theta \end{align*}$

bsdfjnlkasn · « **Reply #1601 on:** March 06, 2017, 06:15:16 pm »

Quote from: kiwiberry on March 06, 2017, 06:10:11 pm

Another method - the question does say to use the t-formulae
We know that if $t=\tan \frac{\theta}{2}$, then
$\tan \theta = \frac{2t}{1-t^2}$
Subbing into the question,
$\begin{align*}\frac{\tan \theta \tan \frac{\theta}{2}}{\tan \theta -\tan \frac{\theta}{2}}&= \frac{\frac{2t}{1-t^2} \times t}{\frac{2t}{1-t^2}-t} \\&=\frac{2t^2}{2t-t(1-t^2)} \\ &=\frac{2t}{1+t^2} \\ &=\sin \theta \end{align*}$

Hey Kiwiberry!

That's how I initially approached the question and got all the way until the second last step - how were you able to cancel the 2t over an open sign?
(Let me know if that question doesn't make sense - thanks for the awesome solution

kiwiberry · « **Reply #1602 on:** March 06, 2017, 06:29:52 pm »

Quote from: bsdfjn;lkasn on March 06, 2017, 06:15:16 pm

Hey Kiwiberry!

That's how I initially approached the question and got all the way until the second last step - how were you able to cancel the 2t over an open sign?
(Let me know if that question doesn't make sense - thanks for the awesome solution

Hey, no worries

do you mean how I turned the t's into $\sin \theta$, or the step before that?

bsdfjnlkasn · « **Reply #1603 on:** March 06, 2017, 06:36:40 pm »

Quote from: kiwiberry on March 06, 2017, 06:29:52 pm

Hey, no worries do you mean how I turned the t's into $\sin \theta$, or the step before that?

the step before that

RuiAce · « **Reply #1604 on:** March 06, 2017, 06:39:25 pm »

Quote from: bsdfjn;lkasn on March 06, 2017, 06:36:40 pm

the step before that

$\frac{2t}{2t-t(1-t^2)}=\frac{2t}{2t-t+t^3}=\frac{2t}{t+t^3}=\frac{2t}{t(1+t^2)}=\frac{2}{1+t^2}$
Just do it step by step if it was too rapid.

Although, if you're wondering "why" we were allowed to cancel out, there's some technicalities behind that which you won't need to worry about for 3U.

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