Question

Use (60° - 45°) = 15° to find the exact value of cos 15º.vaV2 + V6V-V6(b)(c)4(d)4+ V62

Answer 1

Answer;

`B\text{. }\frac{\sqrt[]{2}+\sqrt[]{6}}{4}`

Explanation;

Given that;

`(60^0-45^0)=15^0`

Hence;

`\text{Cos 15}^0=Cos(60^0-45^0)`

According to trigonometry identity;

`\begin{gathered} Cos(60^0-45^0\text{) = Cos60 Cos45 + Sin60Sin45} \\ Cos(60^0-45^0\text{) }=(1)/(2)(\frac{1}{\sqrt[]{2}})+\frac{\sqrt[]{3}}{2}(\frac{1}{\sqrt[]{2}}) \end{gathered}`

Evaluate the result by finding the LCM

`Cos(60^0-45^0\text{) }=\frac{1+\sqrt[]{3}}{2\sqrt[]{2}}`

Rationalize;

`\begin{gathered} Cos(60^0-45^0\text{) }=\frac{1+\sqrt[]{3}}{2\sqrt[]{2}}*\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ Cos(60^0-45^0\text{) }=\frac{\sqrt[]{2}(1+\sqrt[]{3})}{2\cdot2} \\ Cos(60^0-45^0\text{) }=\frac{\sqrt[]{2}+\sqrt[]{6}}{4} \end{gathered}`

Hence the required reusult is;

`\frac{\sqrt[]{2}+\sqrt[]{6}}{4}`