Find the Exact Value tan(4pi)-sin(-(17pi)/6)+cos(-(13pi)/3)

$\tan (4 π) - \sin (- \frac{17 π}{6}) + \cos (- \frac{13 π}{3})$

Simplify each term.

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\tan (0) - \sin (- \frac{17 π}{6}) + \cos (- \frac{13 π}{3})$

The exact value of $\tan (0)$ is $0$ .

$0 - \sin (- \frac{17 π}{6}) + \cos (- \frac{13 π}{3})$

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$0 - \sin (\frac{7 π}{6}) + \cos (- \frac{13 π}{3})$

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 third quadrant.

$0 - - \sin (\frac{π}{6}) + \cos (- \frac{13 π}{3})$

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

$0 - - \frac{1}{2} + \cos (- \frac{13 π}{3})$

Multiply $- - \frac{1}{2}$ .

Multiply $- 1$ by $- 1$ .

$0 + 1 (\frac{1}{2}) + \cos (- \frac{13 π}{3})$

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

$0 + \frac{1}{2} + \cos (- \frac{13 π}{3})$

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$0 + \frac{1}{2} + \cos (\frac{5 π}{3})$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$0 + \frac{1}{2} + \cos (\frac{π}{3})$

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$0 + \frac{1}{2} + \frac{1}{2}$

Combine fractions.

Combine the numerators over the common denominator.

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

Simplify the expression.

Add $1$ and $1$ .

$\frac{2}{2}$

Divide $2$ by $2$ .

$1$

Find the Exact Value tan(4pi)-sin(-(17pi)/6)+cos(-(13pi)/3)

Related Questions

Support LCMGCF.com by disabling your adblocker.