Respuesta :

From the question;

we are given

[tex]\cos \text{ x = }\frac{1}{4},\text{ }\frac{\pi}{2}<\text{ x }<1[/tex]

we are to find

[tex]cos\text{ (}\frac{\pi}{6}\text{ - x)}[/tex]

To find this we will need to simplify

therefore

[tex]\begin{gathered} \sin ce\text{ }\frac{\pi}{6}=\text{ }\frac{180}{6}\text{ =30} \\ \text{hence} \\ \cos (\frac{\pi}{6}\text{ - x) = cos (30 - x)} \end{gathered}[/tex]

applying trigonometric formula

[tex]\cos (A\text{ - B) = cosAcosB + sinAsinB}[/tex]

Hence

[tex]\cos (30\text{ - x) = cos30cosx + sin30sinx}[/tex]

First, we need to find sinx

Starting with cos x = 1/4

we will apply trigonemetric ratio

SOH CAH TOA

Therefore

[tex]\cos \text{ x = }\frac{\text{adjacent}}{\text{hypothenus}}[/tex]

This gives

adjacent = 1

hypothenus = 4

constructing the triangle we get

From triangle

Applying pythagoras rule

[tex]\begin{gathered} \text{hyp}^2=opp^2+adj^2 \\ \text{therefore} \\ 4^2=opp^2+1^2 \\ 16=opp^2\text{ + 1} \\ \text{opp}^2\text{ = 16 - 1} \\ \text{opp}^2\text{ = 15} \\ \text{opp =}\sqrt[]{15} \end{gathered}[/tex]

finding sin x

using trigonometric ratio

[tex]\begin{gathered} \sin \text{ x = }\frac{\text{opposite}}{hypotenus} \\ \text{opp =}\sqrt[]{15},\text{ hypotenus = 4} \\ \text{therefore} \\ \sin \text{ x = }\frac{\sqrt[]{15}}{4} \end{gathered}[/tex]

Finally

Finding

[tex]\cos (\frac{\pi}{6}-x)[/tex]

recall

[tex]\begin{gathered} \cos (\frac{\pi}{6}-x)\text{ = cos(30 - x)} \\ \text{and} \\ \cos (30-x)\text{ = cos30cosx + sin30sinx} \end{gathered}[/tex]

Therefore

appying our values we have

[tex]\begin{gathered} \cos (30\text{ - x) = cos30cosx + sin30sinx} \\ \cos (30\text{ - x) = }\frac{\sqrt[]{3}}{2}\times\frac{1}{4}\text{ + }\frac{1}{2}\times\frac{\sqrt[]{15}}{4} \\ \cos (30\text{ - x) = }\frac{\sqrt[]{3}}{8}\text{ + }\frac{\sqrt[]{15}}{8} \\ \cos (30\text{ - x) = }\frac{\sqrt[]{3}\text{ + }\sqrt[]{15}}{8} \end{gathered}[/tex]

Therefore

The answer is

[tex]\cos (\frac{\pi}{6}-x)\text{ = }\frac{\sqrt[]{3}+\text{ }\sqrt[]{15}}{8}[/tex]

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