---

title: Calculas Scholarship T2W4
tags: Calculus Scholarship
description: Term 2 Week 4
type: pdf
slideOptions:
theme: white

---

###### tags: `Calculus Scholarship`

### Q1 ###

a). $cos(2A + B) = cos2A cosB – sin2A sinB = (2cos^2A – 1) cosB – 2sinA cosA sinB$
let $A = B = \theta$, so we have
$cos(3θ) = 2cos^3θ–cosθ–2sin^2θcosθ$
$=2cos^3θ–cosθ–2(1 – cos^2θ) cosθ = 4cos^3θ – 3cosθ$
$\implies \frac{cos(3\theta)}{4}=cos^3θ – \frac{3}{4}cosθ$

b). let $x = \frac{2}{3}cos(\theta)$, so the equation become:
$27(\frac{2}{3}cos(\theta)^3)-9(\frac{2}{3}cos(\theta))-1=0$
$\implies 8cos^3(\theta)-6cos(\theta)=1\implies cos^3(\theta)-\frac{3}{4}cos(\theta)=\frac{1}{8}$, so from a), we have
$cos(3\theta)=\frac{1}{2} \implies 3\theta=\frac{\pi}{3},\frac{5\pi}{3},\frac{7\pi}{3}\implies\theta=\frac{\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9}$, so
$x=\frac{2}{3}cos(\frac{\pi}{9}), \frac{2}{3}cos(\frac{5\pi}{9}),\frac{2}{3}cos(\frac{7\pi}{9})$

c). As x is root of the cubic equation, by product rule, we have
$\frac{2}{3}cos(\frac{\pi}{9})\cdot \frac{2}{3}cos(\frac{5\pi}{9})\cdot \frac{2}{3}cos(\frac{7\pi}{9}) = \frac{1}{27}$, which means
$cos(\frac{\pi}{9})\cdot cos(\frac{5\pi}{9})\cdot cos(\frac{7\pi}{9}) = \frac{1}{8}$
$\implies cos(\frac{\pi}{9})\cdot cos(\frac{3\pi}{9})\cdot cos(\frac{5\pi}{9})\cdot cos(\frac{7\pi}{9}) = \frac{1}{16}$

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### Q2 ###

a). (i).

Solve for quadratic, we have $x = \frac{1}{\sqrt{2}}$, omit the process here.,

(ii). Using trig-identity, we have $sinx + \sqrt{1-sin^2x}=\sqrt{2}$, so by (i),
$sinx=\frac{1}{\sqrt{2}}\implies x=\frac{\pi}{4}$

b). $\sec x= \frac{1}{\cos x}$ from basic indentity, with $x \ne (n+\frac{1}{2})\pi$

c). $\sec ^2x = \frac{1}{\cos^2x} = \frac{\cos^2x+\sin^2x}{\cos^2x}=1+\tan^2x$

d) let $y = \cos^{-1}x$, so $\cos y = x$. Then using c), we have $\tan^2 y = \sec^2 y -1 =\frac{1}{\cos^2 y} - 1=\frac{1}{x^2}-1$
$\implies \tan y = \sqrt{\frac{1-x^2}{x^2}}=\frac{\sqrt{1-x^2}}{x} = \tan (\cos^{-1}x)$

e) let $\theta = \sin^{-1}x = \tan^{-1}x$, we have $\sin \theta =x$ and $tan \theta = z$. With same angle $\theta$ in a right-angle triagle, we can assume the length of hypotenuse side is 1, then the adjacent side would be $\sqrt{1-x^2}$, then $\tan \theta = \frac{x}{\sqrt{1-x^2}} = z$

**Note**: (d)(e) involve Inverse trig-functions which are too hard for current NCEA standard. You might choose to ignore them for now.

<br/>

### Q3 ###

a). (i).

$\cos \phi = \cos (\frac{\phi}{2}+\frac{\phi}{2})=\cos \frac{\phi}{2}\cos \frac{\phi}{2}-\sin \frac{\phi}{2}\sin \frac{\phi}{2}=\cos^2\frac{\phi}{2}-\sin^2\frac{\phi}{2}$
$=1-\sin^2\frac{\phi}{2}-\sin^2\frac{\phi}{2}=1-2\sin^2\frac{\phi}{2}$

(ii). When $\phi$ is small, $\sin \frac{\phi}{2} \approx \frac{\phi}{2}$ from the assumption. So
$\cos \phi \approx 1-2(\frac{\phi}{2})^2 = 1-\frac{\phi^2}{2}$

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### Q4 ###

(i). $2\sin \frac{\alpha}{2^n} \cos \frac{\alpha}{2^n} = \sin \frac{2\alpha}{2^n} \implies \sin \frac{\alpha}{2^n} \cos \frac{\alpha}{2^n} = \frac{1}{2}\sin \frac{\alpha}{2^{n-1}}$

(ii). $\cos \alpha = 2 \cos^2 \frac{\alpha}{2} - 1 \implies \cos^2 \frac{\alpha}{2} =\frac{1}{2} + \frac{1}{2}\cos\alpha$
$\implies \cos \frac{\alpha}{2} = \sqrt{\frac{1}{2} + \frac{1}{2}\cos\alpha}$

(iii).$\cos \frac{\pi}{8} = \cos \frac{\pi}{4\cdot2} = \sqrt{\frac{1}{2} + \frac{1}{2}\cos\frac{\pi}{4}} = \sqrt{\frac{1}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2}}$
$=\sqrt{\frac{2+\sqrt{2}}{4}}=\frac{\sqrt{2+\sqrt{2}}}{2}$

<br/>

### Q5 ###

(a). Omitted. Factorize and then use Sum to product formular to get product of $\cos$ equals to 0. Then solve for different group of solutions.

(b).
The first task can be assumed that,

$\ln x$ is the angle in a right-triangle with hypotenuse side as 1 and opposite as a, so another side would be $\sqrt{1-a^2}$;
$\ln y$ is the angle in another right-triangle with hypotenuse side as b and adjacent side as 1, so the opposite side would be $\sqrt{b^2 -1}$
We can now get $\tan (\ln x)$ and $\tan (\ln y)$ with relationship above.
Then follow the compound angle formula of $\tan$ function, we can get the expression.

We only show the details of solving equation here:

$\tan (\ln (xy)) \tan (\ln (\frac{x}{y})) = 1$
$\implies \sin (\ln (xy)) \sin (\ln (\frac{x}{y})) = \cos (\ln (xy)) \cos (\ln (\frac{x}{y}))$
$\implies \cos (\ln (xy)) \cos (\ln (\frac{x}{y})) - \sin (\ln (xy)) \sin (\ln (\frac{x}{y})) = 0$
$\implies \cos (\ln (xy) + \ln (\frac{x}{y})) = 0$
$\implies \cos (2\ln (x)) = 0 \implies 1- 2\sin^2 (\ln (x))=0$
$\implies 1 - 2a^2 = 0$
$\implies a = \pm \frac{1}{\sqrt{2}}$

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### Q6 ###

$\frac{\cos \theta}{1+\sin \theta}-\frac{\sin \theta}{1+\cos \theta} = \frac{\cos\theta+\cos^2\theta-\sin\theta - \sin^2\theta}{1 + \sin \theta + \cos \theta + \sin\theta \cos\theta} = \frac{\cos\theta-\sin\theta +\cos^2\theta - \sin^2\theta}{1 + \sin \theta + \cos \theta + \sin\theta \cos\theta}$
$=\frac{(\cos\theta-\sin\theta) +(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}{1 + \sin \theta + \cos \theta + \sin\theta \cos\theta} = \frac{(\cos\theta-\sin\theta)(1+\cos\theta+\sin\theta)}{1 + \sin \theta + \cos \theta + \sin\theta \cos\theta}$
$=\frac{2(\cos\theta-\sin\theta)(1+\cos\theta+\sin\theta)}{2 + 2\sin \theta + 2\cos \theta + 2\sin\theta \cos\theta} = \frac{2(\cos\theta-\sin\theta)(1+\cos\theta+\sin\theta)}{1 + \sin^2\theta + \cos^2 + 2\sin \theta + 2\cos \theta + 2\sin\theta \cos\theta}$
$=\frac{2(\cos\theta-\sin\theta)(1+\cos\theta+\sin\theta)}{(1+\sin\theta + \cos \theta)^2} = \frac{2(\cos\theta-\sin\theta)}{1+\sin\theta + \cos \theta}$