关于倍角公式的降次题!(4sin^t乘以2cost乘以2cost)=2(1-cos4t)

网友回答

4(sint)^2*2cost*2cost
=4(sint)^2*4(cost)^2
=4(2sintcost)^2
=4(sin2t)^2
=4*(1-cos4t)/2
=2(1-cos4t)
======以下答案可供参考======
供参考答案1:
4(sint)^2*2cost*2cost
=4(sint)^2*4(cost)^2
=4(2sintcost)^2
=4(sin2t)^2
=4*(1-cos4t)/2
=2(1-cos4t)
供参考答案2:
是4*sin^2(t)*2cos(t)*2cos(t)吧
=4(2sin(t)cos(t))^2
=4*sin^2(2t)
=4*(1-cos(4t))/2
=2(1-cos(4t))
供参考答案3:
4(sint)^2*2cost*2cost
=4(sint)^2*4(cost)^2
=4(2sintcost)^2
=4(sin2t)^2
=4*(1-cos4t)/2
=2(1-cos4t)