Question

Find the exact value of each expression. Do not use a calculator.

$3\sin {45}^{\circ }-4\tan \frac{\mathrm{\pi }}{6}$

Short answer

The value of the given expression is$\frac{9\sqrt{2}-8\sqrt{3}}{6}$

Step-by-step solution

Step 1. Given

Given trigonometric function

$3\sin {45}^{\circ }-4\tan \frac{\mathrm{\pi }}{6}$

Step 2. Concept

We will use standard formula of $\sin \theta$ and $\cos \theta$at different angles for calculating the final result of given expression.

Step 3. Calculation of sin45∘

$$\begin{gathered} Considerarightangledisoscelestriangle \\ therefore \\ twoanglesare{45}^{\circ } \\ lettwosidesbe1,1 \\ nowapplyingpythogorasweget \\ {c}^2=a^2+b^2 \\ {c}^2=2 \\ c=\sqrt{2} \\ Now \\ Sin\frac{\mathrm{\pi }}{4}=Sin{45}^{\circ }=\frac{opposite}{hypotenuse}=\frac{b}{c}=\frac{1}{\sqrt{2}} \end{gathered}$$

Step 4. Calculation of tanπ6

$$\begin{gathered} Considerarighttrianlgewithangles{30}^{\circ }\&{90}^{\circ } \\ Leta=1,c=2 \\ Applyingpythagorasweget \\ {a}^2+b^2=c^2 \\ 1+b^2=2^2 \\ {b}^2=3 \\ b=\sqrt{3} \\ Now \\ \sin \frac{\mathrm{\pi }}{6}=\sin {30}^{\circ }=\frac{opposite}{hypotenuse}=\frac{a}{c}=\frac{1}{2} \\ \cos \frac{\mathrm{\pi }}{6}=\cos {30}^{\circ }=\frac{base}{hypotenuse}=\frac{b}{c}=\frac{\sqrt{3}}{2} \\ therefore, \\ \tan \frac{\mathrm{\pi }}{6}=\frac{\sin {\displaystyle \frac{\mathrm{\pi }}{6}}}{\cos {\displaystyle \frac{\mathrm{\pi }}{6}}}=\frac{{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{\sqrt{3}}{2}}}=\frac{1}{\sqrt{3}} \end{gathered}$$

Step 5. Final calculation

Substituting the values of trigonometric expression we get

$$\begin{gathered} 3\sin {45}^{\circ }-4\tan \frac{\mathrm{\pi }}{6} \\ =\frac{3}{\sqrt{2}}-4\left(\frac{1}{\sqrt{3}}\right) \\ =\frac{9\sqrt{2}-8\sqrt{3}}{6} \end{gathered}$$