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3 years ago

What is the domain of y = cos θ

Mathematics

2 answers:

ElenaW [278]3 years ago

5 0

All real numbers / all x
if you're asking for range as well then its -1<x<1

rosijanka [135]3 years ago

3 0

Answer:

The domain is all real numbers or $x\in(-\infty,\infty)$

Step-by-step explanation:

The definition of domain is :

Domain is the set of x values for which the function is defined.

The given function is y = cos θ and we know that θ can take any value. In other words, for any value of θ, the function y = cos θ is defined.

Therefore, we can conclude that the domain of y = cos θ is the set of all real values.

In interval notation we can write it as

$x\in(-\infty,\infty)$

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Step-by-step explanation:

Given equation: $x^2-4x+2=0$

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To find the roots, use the quadratic formula:

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Therefore:

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$\begin{aligned}\implies x & =\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)}\\& =\dfrac{4 \pm \sqrt{8}}{2}\\& =\dfrac{4 \pm 2\sqrt{2}}{2}\\& =2 \pm \sqrt{2}\end{aligned}$

$\textsf{Let }a=2+\sqrt{2}$

$\textsf{Let }b=2-\sqrt{2}$

Therefore:

$\begin{aligned}\implies \dfrac{a}{b^2}+\dfrac{b}{a^2} & = \dfrac{2+\sqrt{2}}{(2-\sqrt{2})^2}+\dfrac{2-\sqrt{2}}{(2+\sqrt{2})^2}\\\\& = \dfrac{2+\sqrt{2}}{6-4\sqrt{2}}+\dfrac{2-\sqrt{2}}{6+4\sqrt{2}}\\\\& = \dfrac{(2+\sqrt{2})(6+4\sqrt{2})+(2-\sqrt{2})(6-4\sqrt{2})}{(6-4\sqrt{2})(6+4\sqrt{2})}\\\\& = \dfrac{12+8\sqrt{2}+6\sqrt{2}+8+12-8\sqrt{2}-6\sqrt{2}+8}{36+24\sqrt{2}-24\sqrt{2}-32}\\\\& = \dfrac{40}{4}\\\\& = 10\end{aligned}$