Compute \(4 \cos 50^\circ - \tan 40^\circ\), without using a calculator. How can I do this?

Guest Jun 21, 2020

#1**+3 **

**Compute **

**\(4 \cos(50^\circ) - \tan(40^\circ)\),**

**without using a calculator. **

**How can I do this?**

\(\begin{array}{|rcll|} \hline && \mathbf{4 \cos(50^\circ) - \tan(40^\circ)} \quad | \quad \cos(50^\circ)=\cos(90^\circ-40^\circ)=\sin(40^\circ) \\\\ &=& 4\sin(40^\circ) - \tan(40^\circ) \\\\ &=& 4 \sin(40^\circ)*\dfrac{\cos(40^\circ)}{\cos(40^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{2*2 \sin(40^\circ)\cos(40^\circ)} {\cos(40^\circ)} - \tan(40^\circ) \quad | \quad 2 \sin(40^\circ)\cos(40^\circ) = \sin(80^\circ) \\\\ &=& \dfrac{2\sin(80^\circ)} {\cos(40^\circ)} - \tan(40^\circ) \quad | \quad \sin(80^\circ)= \sin(180^\circ-80^\circ)=\sin(100^\circ) \\\\ &=& \dfrac{2\sin(100^\circ)} {\cos(40^\circ)} - \tan(40^\circ) \quad | \quad \cos(60^\circ)=\dfrac{1}{2} \ \text{or}\ 2=\dfrac{1}{\cos(60^\circ)} \\\\ &=& \dfrac{\sin(100^\circ)} {\cos(40^\circ)\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{\sin(40^\circ+60^\circ)} {\cos(40^\circ)\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{\sin(40^\circ)\cos(60^\circ)+\cos(40^\circ)\sin(60^\circ)} {\cos(40^\circ)\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{\sin(40^\circ)\cos(60^\circ)} {\cos(40^\circ)\cos(60^\circ)}+ \dfrac{\cos(40^\circ)\sin(60^\circ)} {\cos(40^\circ)\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{\sin(40^\circ)} {\cos(40^\circ)} + \dfrac{\sin(60^\circ)} {\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \tan(40^\circ) + \dfrac{\sin(60^\circ)} {\cos(60^\circ)} - \tan(40^\circ) \\\\ &=& \dfrac{\sin(60^\circ)} {\cos(60^\circ)} \quad | \quad \sin(60^\circ)=\dfrac{\sqrt{3}}{2},\ \cos(60^\circ)=\dfrac{1}{2} \\\\ &=& \dfrac{\sqrt{3}}{2} \above 1pt \dfrac{1}{2} \\\\ &=& \mathbf{\sqrt{3}} \\ \hline \end{array}\)

heureka Jun 22, 2020