Question

(a) $\sin \frac{\mathrm{\pi }}{4}·\cos \frac{\pi }{3}=\_\_\_\_\_\_$

(b)$\tan \frac{\mathrm{\pi }}{4}-\sin \frac{\mathrm{\pi }}{6}=\_\_\_\_\_\_\_\_$

Short answer

(a) $\sin \frac{\mathrm{\pi }}{4}·\cos \frac{\pi }{3}=\frac{1}{2\sqrt{2}}$

(b)localid="1646824125334" $\tan \frac{\mathrm{\pi }}{4}-\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}$

Step-by-step solution

Part (a) Step 1. Find the value

We know,

$$\begin{gathered} \sin \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}} \\ \cos \frac{\pi }{3}=\frac{1}{2} \end{gathered}$$

Therefore,

$$\begin{gathered} \sin \frac{\mathrm{\pi }}{4}·\cos \frac{\pi }{3}=\frac{1}{\sqrt{2}}·\frac{1}{2} \\ \sin \frac{\mathrm{\pi }}{4}·\cos \frac{\pi }{3}=\frac{1}{2\sqrt{2}} \end{gathered}$$

Part (b) Step 1. Find the value

We know,

$$\begin{gathered} \tan \frac{\mathrm{\pi }}{4}=1 \\ \sin \frac{\mathrm{\pi }}{6}=\frac{1}{2} \end{gathered}$$

Therefore,

localid="1646824093386" $$\begin{gathered} \tan \frac{\mathrm{\pi }}{4}-\sin \frac{\mathrm{\pi }}{6}=1-\frac{1}{2} \\ \tan \frac{\mathrm{\pi }}{4}-\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2} \end{gathered}$$