Prelim 2016 Maths Help Thread (2 Viewers)

Green Yoda · Apr 17, 2016

help:
prove that
(sin^2(a)*cos^2(b))-(cos^2(a)*sin^2(b))=sin^2(a)-sin^2(b)

InteGrand · Apr 17, 2016

Rathin said:
help:
prove that
(sin^2(a)*cos^2(b))-(cos^2(a)*sin^2(b))=sin^2(a)-sin^2(b)

$\noindent Here's a `lazy' way of doing it.$

$\noindent \textbf{Claim.} Let $b\in \mathbb{R}$ be fixed. Then $\sin^2 x \cos^2 b - \cos^2 x \sin^2 b = \sin^2 x - \sin^2 b$ for all real $x$.$

$\noindent \textbf{Proof.} Note that the LHS and RHS (viewed as functions of $x$) are continuous and differentiable everywhere on $\mathbb{R}$. We have$

$\begin{align*}\frac{ \mathrm{d} }{ \mathrm{d} x} \left( LHS(x) \right) &= 2 \sin x \cos x \cos^2 b + 2\sin x \cos x \sin^2 b \\ &= 2\sin x \cos x \quad (\text{using the Pythagorean trig. identity}) \\ &=\frac{\mathrm{d}}{\mathrm{d} x}\left( RHS(x)\right). \end{align*}$

$\noindent So the LHS and RHS differ by a constant $C$. Since when $x=0$, $LHS(0) = 0 - 1\cdot \sin^2 b = -\sin^2 b$ and $RHS(0) = 0 - \sin^2 b = -\sin^2 b=LHS(0)$, this constant $C$ is 0. So $LHS(x) = RHS(x)$ for all real $x$. This proves the claim.$

Green Yoda · Apr 17, 2016

InteGrand said:
$\noindent Here's a `lazy' way of doing it.$

$\noindent \textbf{Claim.} Let $b\in \mathbb{R}$ be fixed. Then $\sin^2 x \cos^2 b - \cos^2 x \sin^ b = \sin^2 x - \sin^2 b$.$

$\noindent \textbf{Proof.} Note that the LHS and RHS (viewed as functions of $x$) are continuous and differentiable everywhere on $\mathbb{R}$. We have$

$\begin{align*}\frac{ \mathrm{d} }{ \mathrm{d} x} \left( LHS(x) \right) &= 2 \sin x \cos x \cos^2 b + 2\sin x \cos x \sin^2 b \\ &= 2\sin x \cos x \quad (\text{using the Pythagorean trig. identity}) \\ &=\frac{\mathrm{d}}{\mathrm{d} x}\left( RHS(x)\right). \end{align*}$

$\noindent So the LHS and RHS differ by a constant $C$. Since when $x=0$, $LHS(0) = 0 - 1\cdot \sin^2 b = -\sin^2 b$ and $RHS(0) = 0 - \sin^2 b = -\sin^2 b=LHS(0)$, this constant $C$ is 0. So $LHS(x) = RHS(x)$ for all real $x$. This proves the claim.$

bruh this is purely 2u yr 11 trig.......I dunno differentiation yet lol
Can you do it another way?

InteGrand · Apr 17, 2016

Rathin said:
bruh this is purely 2u yr 11 trig.......I dunno differentiation yet lol
Can you do it another way?

Yes. (Sorry, forgot this was in prelim thread and you hadn't learnt calculus yet.)

KingOfActing · Apr 17, 2016

Rathin said:
bruh this is purely 2u yr 11 trig.......I dunno differentiation yet lol
Can you do it another way?

$Hint: Write $ \cos^2{x} $ as $ 1 - \sin^2{x}$ and $\cos^2{b} $ as $ 1 - \sin^2{b}$ then expand.$

InteGrand · Apr 17, 2016

Rathin said:
help:
prove that
(sin^2(a)*cos^2(b))-(cos^2(a)*sin^2(b))=sin^2(a)-sin^2(b)

$\noindent Here's a more typical way of doing it.$

$\noindent Note that by rearranging, the statement is equivalent to $\sin^2 a \cos ^2 b - \cos^2 a \sin^2 b - \sin^2 a + \sin^2 b = 0$. But in this statement, using factorisation and the Pythagorean Trig. Identity, $LHS = \sin^2 a \left(\cos^2 b -1\right) + \sin^2 b \left(1-\cos^2 a\right) = -\sin^2 a \sin^2 b + \sin^2 b \sin^2 a = 0 = RHS$, so this statement is true. Hence the original statement is true. Q.E.D.$

Nailgun · Apr 17, 2016

my pleb working
$\sin ^{ 2 }{ a } =1-\cos ^{ 2 }{ a } \\\\ =\cos ^{ 2 }{ b(1- } \cos ^{ 2 }{ a } )-\cos ^{ 2 }{ a } \sin ^{ 2 }{ b } \\ =\cos ^{ 2 }{ b } -\cos ^{ 2 }{ a } \cos ^{ 2 }{ b } -\cos ^{ 2 }{ a } \sin ^{ 2 }{ b } \\ =\cos ^{ 2 }{ b } -\cos ^{ 2 }{ a } (\cos ^{ 2 }{ b } +\sin ^{ 2 }{ b } )\\ =\cos ^{ 2 }{ b } -\cos ^{ 2 }{ a } \\ =1-\sin ^{ 2 }{ b } -1+\sin ^{ 2 }{ a } \\ =\sin ^{ 2 }{ a } -\sin ^{ 2 }{ b } \\ Q.E.D$

eyeseeyou · Apr 18, 2016

Nailgun said:
my pleb working
$\sin ^{ 2 }{ a } =1-\cos ^{ 2 }{ a } \\\\ =\cos ^{ 2 }{ b(1- } \cos ^{ 2 }{ a } )-\cos ^{ 2 }{ a } \sin ^{ 2 }{ b } \\ =\cos ^{ 2 }{ b } -\cos ^{ 2 }{ a } \cos ^{ 2 }{ b } -\cos ^{ 2 }{ a } \sin ^{ 2 }{ b } \\ =\cos ^{ 2 }{ b } -\cos ^{ 2 }{ a } (\cos ^{ 2 }{ b } +\sin ^{ 2 }{ b } )\\ =\cos ^{ 2 }{ b } -\cos ^{ 2 }{ a } \\ =1-\sin ^{ 2 }{ b } -1+\sin ^{ 2 }{ a } \\ =\sin ^{ 2 }{ a } -\sin ^{ 2 }{ b } \\ Q.E.D$

What does Q.E.D stand for?

jathu123 · Apr 18, 2016

eyeseeyou said:
What does Q.E.D stand for?

In Latin, it translates into quod erat demonstrandum, meaning "which is what had to be proven". It's mainly used in the end of a mathematical proof

InteGrand · Apr 18, 2016

eyeseeyou said:
What does Q.E.D stand for?

QED is an abbreviation of the Latin words "Quod Erat Demonstrandum" which loosely translated means "that which was to be demonstrated". It is usually placed at the end of a mathematical proof to indicate that the proof is complete.

Source: http://www.mathopenref.com/qed.html.

clareh08 · Apr 18, 2016

If in 2014, the Australias population was 23 246 692, with a growth rate of approximately 1.14%, what was the population in 2015?

Please help! What is the formula and how would you work out this question

KingOfActing · Apr 18, 2016

clareh08 said:
If in 2014, the Australias population was 23 246 692, with a growth rate of approximately 1.14%, what was the population in 2015?

Please help! What is the formula and how would you work out this question

Unless there's something I'm missing, isn't it just 23,246,692*(1+0.0114) = 23511704.3, and since you can't have 0.3 of a person, the answer is 23,511,704

$$If a quantity $P$ grows at rate $r$, and was equal to $P_0$ initially, then after $n$ time measurements w.r.t the rate, $P = P_0(1+r)^n$

Here the initial measurement was 23 246 692, and 1 year passed.

Green Yoda · Apr 20, 2016

Help:
prove sin^2(θ)tan(θ)+cos^2(θ)cot(θ)+2sin(θ)cos(θ) = tan(θ)+cot(θ)

My working out so far:
LHS=(1-cos^2(θ))tan(θ)+cos^2(θ)cot(θ)+2sin(θ)cos(θ)
=tan(θ)-sin(θ)cos(θ)+cos^2(θ)cot(θ)+2sin(θ)cos(θ)
=tan(θ)+cos^2(θ)cot(θ)+sin(θ)cos(θ)

parad0xica · Apr 20, 2016

Rathin said:
Help:
prove sin^2(θ)tan(θ)+cos^2(θ)cot(θ)+2sin(θ)cos(θ) = tan(θ)+cot(θ)

My working out so far:
LHS=(1-cos^2(θ))tan(θ)+cos^2(θ)cot(θ)+2sin(θ)cos(θ)
=tan(θ)-sin(θ)cos(θ)+cos^2(θ)cot(θ)+2sin(θ)cos(θ)
=tan(θ)+cos^2(θ)cot(θ)+sin(θ)cos(θ)

cos^2 = 1 - sin^2 and you are done

KingOfActing · Apr 20, 2016

Rathin said:
Help:
prove sin^2(θ)tan(θ)+cos^2(θ)cot(θ)+2sin(θ)cos(θ) = tan(θ)+cot(θ)

My working out so far:
LHS=(1-cos^2(θ))tan(θ)+cos^2(θ)cot(θ)+2sin(θ)cos(θ)
=tan(θ)-sin(θ)cos(θ)+cos^2(θ)cot(θ)+2sin(θ)cos(θ)
=tan(θ)+cos^2(θ)cot(θ)+sin(θ)cos(θ)

You're almost there. Continuing on:

$\begin{align*}\tan{\theta} + (1-\sin^2{\theta})\cot{\theta} + \sin{\theta}\cos{\theta} &= \tan{\theta} + \cot\theta - \sin\theta\cos\theta + \sin\theta\cos\theta \\&=\tan\theta + \cot\theta \end{align*}$

Green Yoda · Apr 22, 2016

help:
express tan(3θ) in terms of θ

InteGrand · Apr 22, 2016

Rathin said:
help:
express tan(3θ) in terms of θ

You mean in terms of tan(θ), right?

$\noindent To do this, apply the compound angle formula for tan to $\tan \left(2\theta + \theta\right)$, and use double angle formula after that to get the $\tan 2\theta$'s in terms of $\tan \theta$.$

Green Yoda · Apr 22, 2016

InteGrand said:
You mean in terms of tan(θ), right?

$\noindent To do this, apply the compound angle formula for tan to $\tan \left(2\theta + \theta\right)$, and use double angle formula after that to get the $\tan 2\theta$'s in terms of $\tan \theta$.$

Thats what I did but the textbook wants it in Just θ....

InteGrand · Apr 22, 2016

Rathin said:
Thats what I did but the textbook wants it in Just θ....

Well it is already in terms of θ.

Green Yoda · Apr 22, 2016

I have the worked our solution and they did some weird shit...
final answer: 3tan(θ)-tan^2(θ)/1-3tan^2(θ)

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