设a为二象限角,tan(a+π/4)=1/2,求sina+cosa的值

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设a为二象限角,tan(a+π/4)=1/2,求sina+cosa的值

解由a为二象限角,tan(a+π/4)=1/2
知a+π/4仍为二象限角,
则由sin²(a+π/4)+cos²(a+π/4)=1
即tan²(a+π/4)+1=1/cos²(a+π/4)
即cos²(a+π/4)=1/(tan²(a+π/4)+1)=1/((1/2)²+1)=4/5
即sin²(a+π/4)=1/5
由a+π/4仍为二象限角,
即sin(a+π/4)=√5/5
即sina+cosa
=√2(√2/2sina+√2/2cosa)
=√2sin(a+π/4)
=√2*√5/5
=√10/5.