已知cos(θ-π/4)=3/5,θ∈(π/2,π),求cosθ的值？
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因为：θ∈(π/2,π),
所以：θ-π/4∈(π/4,3π/4)
因此：sin(θ-π/4)＞0
又：cos(θ-π/4)=3/5
而：sin²(θ-π/4)+cos²(θ-π/4)=1
所以,sin²(θ-π/4)=1-cos²(θ-π/4)=1-(3/5)²=16/25
由此得：sin(θ-π/4)=4/5
cosθ=cos[(θ-π/4)+π/4]
=cos(θ-π/4)cos(π/4)-sin(θ-π/4)sin(π/4)
=(3/5)(√2)/2-(4/5)(√2)/2
=(3/10)√2-(4/10)√2
=-(√2)/10,10,
所以：θ-π/4∈(π/4,3π/4)
因此：sin(θ-π/4)＞0
又：cos(θ-π/4)=3/5
而：sin²(θ-π/4)+cos²(θ-π/4)=1
所以,sin²(θ-π/4)=1-cos²(θ-π/4)=1-(3/5)²=16/25
由此得：sin(θ-π/4)=4/5
cosθ=cos[(θ-π/4)+π/4]
=cos(θ-π/4)cos(π/4)-sin(θ-π/4)sin(π/4)
=(3/5)(√2)/2-(4/5)(√2)/2
=(3/10)√2-(4/10)√2
=-(√2)/10,10,
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