Lena Bell
2022-07-03
Answered

If $\mathrm{tan}x+{\mathrm{tan}}^{2}x+{\mathrm{tan}}^{3}x=1$ then we have to find the value of $2{\mathrm{cos}}^{6}x-2{\mathrm{cos}}^{4}x+{\mathrm{cos}}^{2}x$

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Does the identity ${\mathrm{cos}}^{2}\left(x\right)+{\mathrm{sin}}^{2}\left(x\right)=1$ hold in a unital Banach algebra where 1 is the unit?

Let's assume that we have an unital Banach algebra T and we define sine and cosine using the normal power series definition as for$\mathbb{R}$ . Let $x\in T$ and let 1 be the unit of T. Does the Pythagorean trigonometric identity ${\mathrm{cos}}^{2}\left(x\right)+{\mathrm{sin}}^{2}\left(x\right)=1$ still hold?

Let's assume that we have an unital Banach algebra T and we define sine and cosine using the normal power series definition as for

asked 2022-04-06

I need to solve the following limit without using L'Hopital's rule:

$\underset{x\to 0}{lim}{(1+\mathrm{sin}\left(x\right))}^{\frac{1}{x}}$

The thing is that I can not figure out what to do. One of my ideas was to apply this rule:$a}^{x}={e}^{\mathrm{ln}\left({a}^{x}\right)}={e}^{x\cdot \mathrm{ln}\left(a\right)$ , getting this:

$\underset{x\to 0}{lim}{e}^{\frac{1}{x}\mathrm{ln}(1+\mathrm{sin}\left(x\right))}$

I already know that the answer is e, so the exponent is definitely 1. However, I tried everything I could but have no idea how to solve$\underset{x\to 0}{lim}\left(\frac{1}{x}\mathrm{ln}(1+\mathrm{sin}\left(x\right))\right)$ which needs to be 1.

I would really appreciate your help, and if you find a totally different way to solve the limit without using L'Hospital's rule it will be good as well.

The thing is that I can not figure out what to do. One of my ideas was to apply this rule:

I already know that the answer is e, so the exponent is definitely 1. However, I tried everything I could but have no idea how to solve

I would really appreciate your help, and if you find a totally different way to solve the limit without using L'Hospital's rule it will be good as well.