化简-p^2tan135°+q^2sin90°-2pqcos0°

化简-p^2tan135°+q^2sin90°-2pqcos0° sin²40°+sin²50°+2-tan135°怎么算= sin(-60°)+cos225°+tan135° sin(-60°)+cos225°+tan135° 高一数学三角比的求值题 sin^2 120°+cos90°+tan135°-cos^2(-330°)+sin(-240°) 求仔细的过程 5sin0+2sin90-3cos270+4sin(-30)coa240-tan135 化简下列psinx+qcosπ/2+ktan2π-p²tan135°+q²sin90°-2pqcos0°a²tanπ/4-b²sin3π/2+adcosπ-abtan5π/4mtanπ+ncosπ/2-psinπ-qcos3π/2-rsin2π 2sin90°+tan45°+2cos60°+tan135°等多少 化简 5sin0度+2sin90度-3cos270度-tan135度 tan135°等于多少? (p+2q)(2p-q)-(p+q)(p-q) ①化简1+cos(π/2 +α)*sin(π/2 -α)*tan(π+α)②计算(1)tan(π/5)+tan(2π/5)+tan(3π/5)+tan(4π/5) (2)sin(-60°)+cos(225°)+tan135°(3)cos(π/5)+cos(2π/5)+cos(3π/5)+cos(4π/5)（4）tan10°+tan170°+sin1866°-sin(-606°) 5sin0· +2sin90·_3cos270· +4sin(-30·)cos240·-tan135· 每个数字后的·是度的意思 cosθ-sinθ=p ,cos3θ+sin3θ=q 证明q=3p-2p^3 设函数f(x)=p·q,其中向量p=(sin x,cos x+sin x),q=(2cos x,cos sin x) ( (p-q)^4/(q-p)^3*(p-q)^2 计算 计算:(p-q)^2 * (p-q)^3 * (q-p)^5 (p-q)^5/(q-p)^2·（p-q)^3