How do you know how to solve sec 45?

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How do you know how to solve sec 45?

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Answer 1 · 2018-06-21T18:00:50

$sec (45^{\circ}) = \frac{\sqrt{2}}{1} = \sqrt{2}$ $sec(45^@)=sqrt2/1=sqrt2$

Explanation:

$sec (45^{\circ})$ $sec(45^@)$

$= \frac{1}{cos (45^{\circ})}$ $=1/cos(45^@)$

$sin (45^{\circ}) = cos (45^{\circ})$ $sin(45^@)=cos(45^@)$

we have an isosceles right triangle
where the ratio of its sides are $1 : 1 : \sqrt{2}$ $1:1:sqrt2$

Cosine = adjacent/hypotenuse

1/cosine=hypotenuse/adjacent
thus:

$sec (45^{\circ}) = \frac{\sqrt{2}}{1} = \sqrt{2}$ $sec(45^@)=sqrt2/1=sqrt2$

Answer 2 · 2018-06-22T00:02:08

$sec 45 = \sqrt{2}$ $sec45=sqrt2$

Explanation:

We know that $sec x$ $secx$ is defined as $\frac{1}{cos x}$ $1/cosx$ , so we are essentially being asked to find

$\frac{1}{cos 45}$ $1/cos45$

On the unit circle, the coordinates for $45^{\circ}$ $45^@$ are $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ $(sqrt2/2,sqrt2/2)$ , where the $x$ $x$ -coordinate is the value for cosine.

This means

$\frac{1}{cos 45} = \frac{1}{\frac{\sqrt{2}}{2}}$ $1/cos45=1/(sqrt2/2)$

Which also means that

$sec 45 = \frac{1}{\frac{\sqrt{2}}{2}}$ $sec45=1/(sqrt2/2)$

which simplifies to

$sec 45 = \frac{2}{\sqrt{2}}$ $sec45=2/sqrt2$

The convention is to not have an irrational number in the denominator, so we can multiply the top and bottom by $\sqrt{2}$ $sqrt2$ . We get

$sec 45 = \frac{2 \sqrt{2}}{\sqrt{2} \sqrt{2}}$ $sec45=(2sqrt2)/(sqrt2sqrt2)$

$sec45=(cancel2sqrt2)/(cancel2)$

$sec45=sqrt2$

Hope this helps!

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How do you know how to solve sec 45?

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