1、
f(x)=√2cos(x-兀/12)
所以
f(兀/3)=√2cos(兀/3-兀/12)
=√2cos(兀/4)
=√2 *√2/2
=1
2、
cos⊙=3/5,而⊙∈(3/2兀,2兀)
所以sin⊙= -4/5
于是
f(⊙-兀/6)
=√2cos(⊙-兀/6-兀/12)
=√2cos(⊙-兀/4)
=√2* [cos⊙*cos(兀/4) +sin⊙*sin(兀/4)]
=√2* (3/5 -4/5) *√2/2
= -1/5
1、
f(x)=√2cos(x-兀/12)
所以
f(兀/3)=√2cos(兀/3-兀/12)
=√2cos(兀/4)
=√2 *√2/2
=1
2、
cos⊙=3/5,而⊙∈(3/2兀,2兀)
所以sin⊙= -4/5
于是
f(⊙-兀/6)
=√2cos(⊙-兀/6-兀/12)
=√2cos(⊙-兀/4)
=√2* [cos⊙*cos(兀/4) +sin⊙*sin(兀/4)]
=√2* (3/5 -4/5) *√2/2
= -1/5