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Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favorite Private Tutor



Last Build Date: Sun, 04 Feb 2018 14:07:50 +0000

 



Comment on 2010 GCE ‘O’ Level Exam Timetable – Black Friday by Deepak Tiwari

Sun, 04 Feb 2018 14:07:50 +0000

Great Peace of article I like it. I did my schooling in 2005 and it really feels great when I listen to these topics.



Comment on Trigonometry Type R by Miss Loi

Sat, 30 Apr 2016 12:21:43 +0000

For the second part, we simply expand the modulus function as follows [tex] \begin{flalign*} 8\cos x-15\sin x \right ) &= 6; \quad 8\cos x-15\sin x \right ) = -6 &\\ 17\cos \left ( x+1.08 \right ) &= 6; \quad 17\cos \left ( x+1.08 \right ) = -6 &\\ \cos \left ( x+1.08 \right ) &= \frac{6}{17}; \quad \cos \left ( x+1.08 \right ) = -\frac{6}{17} & \end{flalign*} \therefore \text{basic angle } \alpha = \cos^{-1} \frac{6}{17} = 1.210 \\ [/tex] Using the SATC quadrants, we have for cos−1 (6/17): [tex] \begin{align*} x+1.08 &= 1.210, 2\pi - 1.210 \\ x &= 0.130, 3.993 \\ &\approx 0.13, 3.99 \end{align*} [/tex] and for cos−1 (−6/17) [tex] \begin{align*} x+1.08 &= \pi - 1.210, \pi + 1.210 \\ x &= 0.8515, 3.271 \\ &\approx 0.852, 3.27 \end{align*} [/tex] [tex]\therefore x \approx 0.13, 0.852, 3.27, 3.99 \text{ rad (3sf)}[/tex]



Comment on Trigonometry Type R by Miss Loi

Sat, 30 Apr 2016 12:21:03 +0000

Look ma! A math question in the comments section! ^^ So you have the R-formula [tex]8\cos x-15\sin x = 17\cos \left ( x+1.08 \right )[/tex] And from the notes above, the maximum/minimum value of a R-formula is ±R so [tex] \begin{alignat*}{3} -17 &\leq 17\cos \left ( x+1.08 \right ) &&\leq 17 \\ -17 &\leq 8\cos x-15\sin x &&\leq 17 \end{alignat*} [/tex] At this juncture, there'll be a insatiable urge to flip the centre term and make the careless mistake of flipping the inequality signs, +20 on both sides to get what you want i.e. [tex] {\color{highlight} \begin{alignat*}{3} -17 &\geq & -\left (8\cos x-15\sin x \right ) &&\geq 17 \\ 3 &\geq & 20 - 8\cos x+15\sin x &&\geq 37 \end{alignat*} \tag{✘}} [/tex] Not only does the above don't make any mathematical sense (3 ≥ x ≥ 37!), one has to always consider the nature of the graph you're dealing with when it comes to inequalities. From your Trigonometric Equations & Graphs, you know that the max/min values of a A cos x curve is still ±A, even when you flip it around for −A cos θ, as shown: (image) So you may now 安安心心 proceed in a as follows: [tex] \begin{alignat*}{3} -17 &\leq &-\left (8\cos x-15\sin x \right ) &\leq 17 \\ 3 &\leq 20&-\left (8\cos x-15\sin x \right ) &\leq 37 \\ 3 &\leq 20&-17\cos \left ( x+1.08 \right ) &\leq 37 \end{alignat*} [/tex] ∴ Maximum value of 37 occurs when [tex] \begin{align*} 20-17\cos \left ( x+1.08 \right ) &= 37 \\ 17\cos \left ( x+1.08 \right ) &= -17 \\ \cos \left ( x+1.08 \right ) &= -1 \\ x+1.08 &= \pi \\ x &= \pi - 1.08 \\ &\approx 2.06 \, \text{rad (3sf)} \end{align*} [/tex] and minimum value of 3 occurs when [tex] \begin{align*} 20-17\cos \left ( x+1.08 \right ) &= 3 \\ -17\cos \left ( x+1.08 \right ) &= -17 \\ \cos \left ( x+1.08 \right ) &= 1 \\ x+1.08 &= 0, 2\pi \\ x &= -1.08 \text{(rej.)}, 2\pi - 1.08 \\ &\approx 5.20 \, \text{rad (3sf)} \end{align*} [/tex] Note: x = −1.08 rejected ∵ 0 ≤ x ≤ 2π



Comment on Trigonometry Type R by Agnes

Thu, 28 Apr 2016 05:56:46 +0000

Hi. I was wondering if you can help me with these questions. 8 cos x - 15 sin x = 17 cos (x + 1.08) -------- (R formula) 1. Find the minimum and maximum value of 20 - 8 cos x + 15 sin x and the corresponding x values for 0 ≤ x ≤ 2π. 2. Solve the equation |8 cos x - 15 sin x| = 6 for 0 ≤ x ≤ 2π. Thanks in advance!






Comment on Trigonometry – You’re An A1 Student If You Can Prove This! by Patrick

Sun, 17 May 2015 05:12:46 +0000

[tex] \begin{align*} \text{LHS} &= \frac{\tan x + \sec x -1}{\tan x - \sec x +1}\\ &= \frac{\tan x + \sec x -[\sec^{2}x - \tan^{2}x]}{\tan x - \sec x +1}\\ &= \frac{(\sec x + \tan x)[1-(\sec x - \tan x)]}{\tan x - \sec x + 1}\\ &= \text{RHS} \end{align*} [/tex]



Comment on Trigonometry – You’re An A1 Student If You Can Prove This! by whatamidoingwithlife

Fri, 24 Apr 2015 14:18:54 +0000

Start from RHS multiply numerator and denominator by tan x - sec x + 1. Expand the numerator (sec x + tan x)(tan x - sec x+1) and identify that (tan^2x-sec^2x)=-1 and you will prove this identity within 30 seconds or so. Sorry for the reply to the really old post — I was stuck with PSLE math at that point in time and I really think this question is a good reminder that there are multiple solutions. The complicated side may not be the easier side and a tinge of creativity is needed. Laziness for the win! ~Some kid with no life <3






Comment on Almost Halloween: GCE O Level Oct/Nov 2014 A-Maths 4047 Paper 2 Suggested Answers & Solutions by Anonymous

Thu, 13 Nov 2014 13:55:57 +0000

Will you be able to post the answers to paper 1 of pure biology?