Using the monotonicity of tangent function, compare the values of two functions in the following groups.

(1)tan(-π/5) and tan (-3π/7): (2) tan1519 and tan 1493.

(3)tan(6 9/π 1 1) and Tan (-53/π1/); (4)tan(7π/8) and tan π

Solution: (1)tanx in (-? π,? π ~ is monotonically increasing, and the first two problems are in this range, -π/5 >-3π/7 (-π/5 is close to the origin on the negative semi-axis, which is large), so Tan (-π/5) > Tan (-3π/7).

(2) First, the period is 180, tan1519 = tan (180× 8+79) = tan 79, tan1493 = tan (/kloc-)

(3) remove the period, tan (69/π11) = tan (9/π11) > 0, tan (-53/π1) =

(4) tan π = 0, tan (7 π/8) > 0, and the first one is larger.

This comparison problem is not as intuitive as drawing a picture first and then looking for coordinates.