If sec ø = 5/4 and 0º<ø<90º how do you find sec 2ø ?

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If sec ø = 5/4 and 0º<ø<90º how do you find sec 2ø ?

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Answer 1 · 2018-02-27T00:57:31

$sec (2 θ) = \frac{25}{7}$ $sec(2theta) = 25/7$ for $sec (θ) = \frac{5}{4}$ $sec(theta) = 5/4$

Explanation:

$sec (θ) = \frac{5}{4}$ $sec(theta) = 5/4$

Recall that $sec θ = \frac{1}{cos θ}$ $sec theta = 1/cos theta$ . Because of this, we can simply take the reciprocal of both sides so we can work with functions we're more used to seeing.

$cos (θ) = \frac{4}{5}$ $color(blue)(cos(theta) = 4/5)$

We are looking for $sec (2 θ)$ $color(green)(sec(2theta))$ , which can also be written in terms of trigonometric functions we are more familiar with.

$sec (2 θ)$ $color(green)(sec(2theta))$

$= \frac{1}{cos (2 θ)}$ $= 1/cos(2theta)$

The double angle identity for cosine states that $cos (2 θ) = 2 {cos}^{2} (θ) - 1$ $color(red)(cos(2theta) = 2cos^2(theta) - 1)$ .

$= \frac{1}{2 {cos}^{2} (θ) - 1}$ $= 1/color(red)(2cos^2(theta) - 1)$

Interestingly, this means that we don't actually have to solve for $θ$ $theta$ to find the value of $sec (2 θ)$ $color(green)(sec(2theta))$ .

$= \frac{1}{2 {(cos (θ))}^{2} - 1}$ $= 1/(2(color(blue)cos(theta))^2 - 1)$

$= \frac{1}{2 {(\frac{4}{5})}^{2} - 1}$ $= 1/(2(color(blue)(4/5))^2 - 1)$

$= \frac{1}{\frac{32}{25} - 1}$ $= 1/(32/25 - 1)$

$= \frac{1}{\frac{7}{25}}$ $= 1/(7/25)$

$= \frac{25}{7}$ $= 25/7$

$therefore$ $sec(2theta) = 25/7$ for $sec(theta) = 5/4$ .

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If sec ø = 5/4 and 0º<ø<90º how do you find sec 2ø ?

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