Question

急求三角函数一题的解法，高手请进

求：cos4X-sin4X-2sinXcosX的值
麻烦过程详细点，谢谢~
那个4不是4次方啊，是4个X

Answer 1

原式=（cos²x+sin²x)(cos²x-sin²x)-2sinxcosx
=1×cos2x-sin2x
=√2(cos2x×√2/2-sin2x×√2/2)
=√2(cos2xcosπ/4-sin2xsinπ/4)
=√2cos(2x+π/4)

Answer 2

4是4次方吗
原式=(cos²x+sin²x)(cos²x-sin²x)-sin2x
=cos²x-sin²x-sin2x
=cos2x-sin2x
=-√2(√2/2*sin2x-√2/2cos2x)
=-√2(sin2xcosπ/4-cos2xsinπ/4)
=-√2sin(2x-π/4)

Answer 3

(cosx)^4－2sinxcosx－(sinx)^4
=(cosx)^4－(sinx)^4－2sinxcosx
=[(cosx)^2+(sinx)^2][(cosx)^2-(sinx)^2]-2sinxcosx
=1*cos2x-sin2x
=√2(cos2x*√2/2-sin2x*√2/2)
=√2(cos2xcosπ/4-sin2xsinπ/4)
=√2cos(2x+π/4)