Question

已知函数F(x)=tan(2x 派/4).设a属于(0,派/4),若F(a/2)=2cos2a,求a的大小

Answer 1

1. 最小正周期T=π/2定义域为2x+π/4∈(2kπ-π/2, 2kπ+π/2)x∈(kπ-3π/8, kπ+π/8)
2. f(α/2)=tan(α+π/4)
=[tanα+tan(π/4)]/[1-tanα*tan(π/4)]
=(tanα+1)(1-tanα)
=(sinα+cosα)/(cosα-sinα)
=2cos2α=2(cos�0�5α-sin�0�5α)
=2(cosα+sinα)(cosα-sinα)
因α（0，四分之π），所以cosα+sinα>0
所以2(cosα-sinα)�0�5
=1cosα-sinα=±√2/2√2sin(π/4-α)
=±√2/2sin(π/4-α)
=±1/2π/4-α
=±π/6解得α=5π/12(舍去)或π/12
∴α=π/12

Answer 2