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Is there someone who can finish Q6iii. Seems like a lot of mental sweat is required.

Q4 Q for reference \tan{x}+2\tan{2x}+4\tan{4x}+...+2^{n-1}\tan2^{n-1}x=\cot{x}-2^{n}\cot\left(2^{n}x\right) for n\geq{1} Start it off let n=1 \tan{x}=\cot{x}-2\cot{2x} \tan{x}=\frac{1}{\tan{x}}-\frac{2}{\tan{2x}} \tan{x}=\frac{\tan{2x}-2\tan{x}}{\tan{x}\tan{2x}}...

1. If you are referring to the sum of digits in 100 then it is 1 and 1 to the power of anything is 1. 2. Product of the digits in 100 is 0 and 0 the power of anything is 0. 3. 97 can you find anything bigger than that.

Once you listened to everyone you will have \frac{8\sqrt{3}-9}{6} and then finish it off. (QED)

Q6i \sin\left(A+B\right)-\sin\left(A-B\right)=\sin{A}\cos{B}+\cos{A}\sin{B}-\left(\sin{A}\cos{B}-\cos{A}\sin{B}\right) That is just 2\cos{A}\sin{B} ii Q for reference \cos{x} + \cos{2x} + \cos{3x} + ... + \cos{nx} =...

Is there a function from LaTeX where I can cross out sections of the working through simplification?

Q5 You start off with \frac{\cos{y}-\cos\left(y+2x\right)}{2\sin{x}}=\sin\left(y+x\right) What you have to do here is that since there are no obvious inroads into the Q you have to expand the \cos\left(y+2x\right) term. Thus, it becomes...

AM-GM inequality as @Octavius said.

thsconline.com is your best friend for that.

Does anyone have some good places? Because I know the rigorous ones are around the Inner West and Parramatta.

Part iii well this can be solved very easily. Draw up a probability table with the following (1,1), (1,2), (1,3), (1,4) all the way till you get (6,6). Once you have that do the sums for each of them. Finish that remove all those sums with a total value of less than 5 and then you will find...

I think we have seen this Q being asked many times on Bored of Studies. I recall this is the third time. Anyways, start with this 1. \left(\sqrt{a}-\sqrt{b}\right)^{2}>0 because a, b > 0. Next, 2. \left(\sqrt{c}-\sqrt{d}\right)^{2}>0 because c, d > 0. Expanding 1. and 2. we will have...

For inequalities, one of the good strategies would be to move everything onto one side and then factorise the problem.

So you're on the n=k+1 step. Alright. Start with this \sum_{r=1}^{k}\left(r+1\right)\times{2^{r}}=n\times{2^{n+1}} For the n=k+1 case \sum_{r=1}^{k+1}\left(r+1\right)\times{2^{r}}=n\times{2^{n+1}} Notice that for the n=k case \sum_{r=1}^{k}\left(r+1\right)\times{2^{r}}=k\times{2^{k+1}}...

For this problem there is a very clever way of approaching this Q focus on i^{1}+i^{2!}+i^{3!}+i^{4!}+... focus on the first 4 terms because you will immediately get I-1-1+1+1 till you get to the 2020 factorial power because every power coming after 4! has a multiple of 4 note that there are...

Use the complement to your advantage.

Fellas this question \sqrt[5]{x^{2}}-4=x^{\frac{2}{5}}-4 Differentiate that and you will have \frac{2}{5x^{\frac{3}{5}}} Put x=-1 and you will have \frac{2}{5} There, we have y+3=-\frac{2}{5}\left(x+1\right). There is a calculation error from \sqrt[5]{x^{2}}-4=x^{\frac{2}{5}}-4 when you subbed...

It's called I want my child to be the best in the state regardless of how much money it takes. So simply these parents are the Whatever it takes type from Captain America from Avengers Endgame.

iPad Pro I mean Notability is a wise choice for writing. That is my take but I think other people's opinions are just as valuable.

Let me give you a clue. Have you tried proving that \sqrt{n} is irrational through contradiction where n is a non-square number? Use that and then add some random rational number