4.1. 二角和差公式 4.2三角和公式 4.3 积化和差公式：4.3.1 口诀法：4.3.2 帅哥记忆法：4...
我们有从x轴正方向逆时针旋转到OB的角∠X1OB为∠B,从x轴正方向逆时针旋转到OA的角∠X1OA为∠A,则∠BOA=∠A-∠B 于是我们有B(cosB,sinB),A(cosA,sinA) 于是AB^{2}=(cosB-cosA)^{2}+(sinB-sinA)^{2} =cos^{2}B-2cosBcosA+cos^{2}A+sin^{2}B-2sinBsinA+sin^{2}A =2-2(cosBcosA+...
我们知道sin(A-B) = sinAcosB - cosAsinB,cos(A-B) = cosAcosB + sinAsinB。 利用这个公式,我们可以得到: sin15° = sin(45°-30°) = sin45°cos30° - cos45°sin30° = 22×32−22×12=6−24\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \times ...
$\therefore $由sinB+sin ( (C-A) )=2sin2A得 $sin\left ( {A+C} \right )+sin\left ( {C-A} \right )=2sin2A$, $\therefore sinAcosC+cosAsinC+sinCcosA-cosCsinA=4sinAcosA$, $\therefore $$cosAsinC=2sinAcosA$, 若$cosA=0$,则$A=\dfrac {\pi } {2}$, $\because $$...
13解:1 ∵cosB=4/5 B为三角形的内角, ∴sinB=3/5∵(AB)/(sinC)=(AC)/(sinB) ∴(AB)/((√2)/2)=-6/5,即 AB=5√2(2) cosA=-cos(C+B)=sinBsinC-cosBcosC∴cosA=-(√2)/(10)又∵A为三角形的内∴cos(A-π/6)=(√3)/2cosA+1/2sinA=(7√2-√6)/(20) 20 结果三...
[B] = Zrotate, Gamma [G] = Yrotate//Sinus Alfa = SinA, cosinus Alfa = cosA. and so on...//First calculate sinus and cosinus for each rotation:sincos(mPtrCtrlState->c3dBodyRot.x+mTotalXBal1, &sinG4, &cosG4);sincos(mPtrCtrlState->c3dBodyRot.z+mTotalZBal1, &sinB4, &cosB...
RHS= m^2+ n^2- 2 = ( cosA + sinB)^2+ ( sinA + cosB)^2 = cos^2A + 2cosAsinB + sin^2B +sin^2A +2sinAcosB + cos^2B -2 =(sin^2A+cos^2A )+ (sin^2B+cos^2B)-2 +2cosAsinB +2sinAcosB = 1+1-2 + 2(cosAsinB +sinAcosB ) =2sin(A+B) =LHS ...
x1x2+y1y2意味着把两个向量对应的分量相乘，再把积相加；|a||b|cosθ意味着把其中一个向量在...
(1-cosA)/sinA=sinA/(1+cosA)和差化积 2sinAcosB=sin(A+B)+sin(A-B) 2cosAsinB=sin(A+B)-sin(A-B) ) 2cosAcosB=cos(A+B)+cos(A-B) -2sinAsinB=cos(A+B)-cos(A-B) sinA+sinB=2sin((A+B)/2)cos((A-B)/2 cosA+cosB=2cos((A+B)/2)sin((A-B)/2) tanA+...
(1+cosA)和差化积 2sinAcosB=sin(A+B)+sin(A-B) 2cosAsinB=sin(A+B)-sin(A-B) ) 2cosAcosB=cos(A+B)+cos(A-B) -2sinAsinB=cos(A+B)-cos(A-B) sinA+sinB=2sin((A+B)/2)cos((A-B)/2 cosA+cosB=2cos((A+B)/2)sin((A-B)/2) tanA+tanB=sin(A+B)/cosAcosB s...