2.α∈(π/2+2kπ, π+2kπ), then α/2 ∈ (π/4+kπ, π/2+kπ) = (π/4+2kπ, π/2+2kπ)∩(5π/4.
3. Because cost+COS (180-t) = 0, COS 90 = 0, COS 180 =-1, the original formula value is-1 (similar to arithmetic progression's idea of summation, it appears in pairs). In fact, if the last term is not cos 180, it can still be summed. Using the product sum and difference formula, multiply sinα first, and then find α. Since sinαcosβ=[sin(β+α)-sin(β-α)]/2, to make the sum of sinαcosβ and sin α cos (β+ 1) cancel one term, then β-α = β+ 1+α, so α =-0.5. So the original formula = [sin (1-0.5)-sin (180+0.5)]/[2 * sin (-0.5)] =-1.
4. Because sin(3π+β)=0(lg 1=0), sinβ=0, cosβ= 1 or-1({cos(π-β)- 1} If it is the denominator, COS β = Otherwise, the original formula = 1+cosβ=2 or 0.