MaxWong

Because $\dfrac13 + \dfrac1{3^2} + \dfrac1{3^3} + \cdots = \dfrac{\dfrac13}{1 - \dfrac13} = \dfrac12$,

$\dfrac12 = 0.\overline{1}_3$

By a similar approach,

$\dfrac17 = 0.\overline{1}_8$

and 

$\dfrac1{b - 1} = 0.\overline{1}_b$

$x = 2^4 \cdot 7^4\\ \sqrt{x} = \sqrt{2^4 \cdot 7^4} = 2^{4/2} \cdot 7^{4/2} = 4\cdot 49 = 196$

Hello Guest:

Is any angle given? I mean, are there any right angles?

If yes, then you don't need sin, cos, or tan. You can calculate it directly by Pythagorean theorem.

By combinatorics, if we want to get a term in x^98, we need to choose 98 x's from all these brackets, and for the remaining two brackets, we choose its constant term.

Therefore, the coefficient of x^98 is 

$\displaystyle \sum_{\substack{1 \leqslant i < j \leqslant 100\\i\in \mathbb Z\\j\in\mathbb Z}} ij$

This sum is $\dfrac{(1 + 2 +\cdots + 100)^2 - (1^2 + 2^2 + \cdots + 100^2)}2$ (why?), which evaluates to 12,582,075.

I believe you meant $\sin(x), \cos(x)\text{, and }\tan(x)$.

We have a mnemonic for that, namely "SOHCAHTOA".

In the following diagram:

AB is opposite to x, so we call it "opposite side"

AC is adjacent to x, so we call it "adjacent side".

BC is the "hypotenuse".

The mnemonic "SOHCAHTOA" helps us remember the following relations:

$\color{red}\text{s}\color{black}\text{in}\;x = \dfrac{\color{red}\text{o}\color{black}\text{pposite side}}{\color{red}\text{h}\color{black}\text{ypotenuse}}$

$\color{red}\text{c}\color{black}\text{os}\;x = \dfrac{\color{red}\text{a}\color{black}\text{djacent side}}{\color{red}\text{h}\color{black}\text{ypotenuse}}$

$\color{red}\text{t}\color{black}\text{an}\;x = \dfrac{\color{red}\text{o}\color{black}\text{pposite side}}{\color{red}\text{a}\color{black}\text{djacent side}}$

The idea:

By law of indices, $(2x)^3 = 2^3 x^3 = 8x^3$

The denominator is 8x³ and the numerator is 2x³. 

Now it is obvious that we can rewrite 8x³ as 4(2x³) and cross out 2x³.

Simplifying gives 1/4.

Notice that $\dfrac1{\log_ab} = \log_ba$ and $c^{\log_c a} = a$.

We will call the first equation (1) and the second equation (2).

$\quad 81^{1/\log_53} + 27^{\log_936}+3^{4/\log_79}\\ \stackrel{(1)}{=} 3^{4\log_35} + 9^{1.5\log_936} + 9^{2\log_97}\\ \stackrel{(2)}{=} 5^4 + 36^{1.5} + 7^2$

The rest is simple.

You can first simplify the expression before substituting x = 2007.

Hint: The expression is constant for any values of x. That means no matter what x is, it always gives the same value.

You can go to that previous post and see the explanation.

By sum-to-product formula:

$\sin \dfrac x2 + \sin \dfrac x3 = 2 \sin \dfrac{5x}{12} \cos\dfrac{x}{12}$

The function $g(x) = \sin \dfrac{5x}{12}$ has period $\dfrac{24\pi}5$, which means $g\left(\dfrac{24n\pi}5 + x\right) = g(x), n\in \mathbb Z$

The function $h(x) = \cos \dfrac{x}{12}$ has period 24 pi, which means $h(24n\pi + x) = h(x), n\in \mathbb Z$

Therefore $f(24\pi + x) = 2g(24\pi + x) h(24\pi + x) = 2g\left(\dfrac{24\pi}5 \cdot 5 + x\right)h(24\pi \cdot 1 + x) = 2g(x) h(x) = f(x)$

Therefore the period is $24\pi$.

No smaller T satisfies $f(T + x) = f(x)$