Trigonometry Examples

$\cos (π) \sin (\frac{7 π}{4}) - \tan (\frac{11 π}{6})$

Step 1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$- \cos (0) \sin (\frac{7 π}{4}) - \tan (\frac{11 π}{6})$

Step 2

The exact value of $\cos (0)$ is $1$ .

$- 1 \cdot 1 \sin (\frac{7 π}{4}) - \tan (\frac{11 π}{6})$

Step 3

Multiply $- 1$ by $1$ .

$- 1 \sin (\frac{7 π}{4}) - \tan (\frac{11 π}{6})$

Step 4

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$- 1 (- \sin (\frac{π}{4})) - \tan (\frac{11 π}{6})$

Step 5

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

$- 1 (- \frac{\sqrt{2}}{2}) - \tan (\frac{11 π}{6})$

Step 6

Multiply $- 1 (- \frac{\sqrt{2}}{2})$ .

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Step 6.1

Multiply $- 1$ by $- 1$ .

$1 \frac{\sqrt{2}}{2} - \tan (\frac{11 π}{6})$

Step 6.2

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

$\frac{\sqrt{2}}{2} - \tan (\frac{11 π}{6})$

Step 7

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the fourth quadrant.

$\frac{\sqrt{2}}{2} - - \tan (\frac{π}{6})$

Step 8

The exact value of $\tan (\frac{π}{6})$ is $\frac{\sqrt{3}}{3}$ .

$\frac{\sqrt{2}}{2} - - \frac{\sqrt{3}}{3}$

Step 9

Multiply $- - \frac{\sqrt{3}}{3}$ .

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Step 9.1

Multiply $- 1$ by $- 1$ .

$\frac{\sqrt{2}}{2} + 1 \frac{\sqrt{3}}{3}$

Step 9.2

Multiply $\frac{\sqrt{3}}{3}$ by $1$ .

$\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3}$

Step 10

The result can be shown in multiple forms.

Exact Form:

$\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3}$

Decimal Form:

$1.28445705 \dots$