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]
