Precalculus Examples

$\sin (\frac{π}{6}) \cos (\frac{π}{4}) - \tan (\frac{x}{4})$

Step 1

Simplify each term.

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Step 1.1

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$\frac{1}{2} \cos (\frac{π}{4}) - \tan (\frac{x}{4})$

Step 1.2

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$\frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \tan (\frac{x}{4})$

Step 1.3

Multiply $\frac{1}{2} \cdot \frac{\sqrt{2}}{2}$ .

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Step 1.3.1

Multiply $\frac{1}{2}$ by $\frac{\sqrt{2}}{2}$ .

$\frac{\sqrt{2}}{2 \cdot 2} - \tan (\frac{x}{4})$

Step 1.3.2

Multiply $2$ by $2$ .

$\frac{\sqrt{2}}{4} - \tan (\frac{x}{4})$

Step 1.4

Rewrite $\tan (\frac{x}{4})$ in terms of sines and cosines.

$\frac{\sqrt{2}}{4} - \frac{\sin (\frac{x}{4})}{\cos (\frac{x}{4})}$

Step 2

Convert from $\frac{\sin (\frac{x}{4})}{\cos (\frac{x}{4})}$ to $\tan (\frac{x}{4})$ .

$\frac{\sqrt{2}}{4} - \tan (\frac{x}{4})$