![已知函數fx=sin(2x+π6)-cos(2x+π3)+cos2x,①f(π12)的值②函數fx單調遞增區間③函數fx最大值和相應x的值,第1張 已知函數fx=sin(2x+π6)-cos(2x+π3)+cos2x,①f(π12)的值②函數fx單調遞增區間③函數fx最大值和相應x的值,第1張](data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==)
已知函數fx=sin(2x π/6)-cos(2x π/3) cos2x,①f(π/12)的值②函數fx單調遞增區間③函數fx最大值和相應x的值網友廻答:
- (1)f(x)=sin(2x π/6)-cos(2x π/3) 2cos²x
= sin2x cosπ/6 cos2x sinπ/6-[ cos2x cosπ/3- sin2x sinπ/3] 2cos²x
=√3 sin2x 2cos²x=√3 sin2x 1 cos2x
=2 sin(2x π/6) 1,
f(π/12)=2sin(2*π/12 π/6) 1=√3 1.
(2)儅2x π/6 ∈[ -π/2 2kπ ,π/2 2kπ ] 函數fx單調遞增
(3)f(x)的最大值是3,
此時2x π/6=2kπ π/2,x=kπ π/6,k∈Z.再教我一條好不好。。你先採納吧!然後發過來我看看!!!已知cos(α β)=5/13,cos(π/2 β)=-3/5,α,β均爲銳角,cosα的值拜托了。。?。。已知cos(α β)=5/13,cos(π/2 β)=-3/5,α,β均爲銳角,cosα的值cos(α β)=5/13所以sin(α β)=12/13cos(π/2 β)=-3/5 ∴sin β=3/5cos β=4/5cos a=cos [(a β)-β]=cos(a β)cosβ sin(a β) sinβ=5/13*4/5 12/13*3/5=56/65嗯嗯嗯謝啦
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