Compare the size, Tan (3 schools / 2 + 1), Tan (3 schools / 2-1)

Compare the size, Tan (3 schools / 2 + 1), Tan (3 schools / 2-1)

Because 1

Compare the sizes of Tan 1, Tan 2 and Tan 3 and prove them

From the function graph of TaNx, we know that TaNx increases monotonically between π / 23 π / 2
tan1=tan（1+π）=tan4.14
π / 2

The numbers of the following groups were compared: (1) tan2 π / 5 and tan3 π / 5 (2) tan2 and tan9 (3) Log1 / 2tan70 ° and Log1 / 2sin25 ° And (1 / 2) cos25 degree Where 1 / 2 of Log1 / 2 is the base and COS 25 ° is the exponent

One
π/3

The order of Tan 1, Tan 2, Tan 3 is______ .

∵1＜π
2＜2＜3＜π
According to the properties of the tangent function, we can get that y = TaNx in (π)
2, π) monotonically increasing
∴tan2＜tan3＜0，tan1＞0
tan1＞tan3＞tan2
So the answer is: TAN1 > tan3 > tan2

How to simplify Tan's 7 / 12 school

tan(7π/12)
=tan(π/2+π/12)
=-cot(π/12)
=-[1+cos(π/6)]/sin(π/6)
=-(1+√3/2)/(1/2)
=-(2+√3)

Simplification: (√ 3 tan12 - 3) / {[4 (cos12) ^ 2 - 2] sin12} (angle system, where √ is the root sign and ^ is the power)

Application thought: reduce the power, cut the string
The original formula = (√ 3sin12 / cos12-3) / {sin12}
=(√3sin12-3cos12)/(2cos24sin12)
=√3(sin12-√3cos12)/(sin24cos24 )
=-2√3sin48/(sin48/2)
=-4√3

How to simplify tan57 ° - tan12 ° - tan57 ° tan12 °/

Analysis:
Because tan45 ° = Tan (57 ° - 12 °) = (tan57 ° - tan12 °) / (1 + tan57 ° tan12 °) = 1
Therefore, tan57 ° - tan12 ° = 1 + tan57 ° tan12 °
The results show that: tan57 ° - tan12 ° - tan57 ° tan12 ° = 1

Compare the sizes of sin 2 / 5 π, cos 6 / 5 π, Tan 7 / 5 π

7π/5>5π/4
tan5π/4=1
So tan 7 π / 5 > 1
Zero

Compare the size of Tan 2 π / 7 and Tan 10 π / 7

tan10π/7=tan(π+3π/7)=tan(3π/7)
∵ y = TaNx is an increasing function on (0, π / 2)
2 π / 7

If tan2 α = - 7.12, Tan α =? α =? α =? α =? α =? α =? α =? α =? α =? α =? α =? α =? α =? α =? α =? α =? α =? α =? α =? α?

tan2α = 2tanα/(1-tan²α) = -7.12
tan²α - 2tanα/7.12 - 1 = 0
tanα = 1/7.12 ± √(1/7.12² +1)
= 0.14 ± 1.06
α = arctan1.2 or α = arctan0.92