题目内容
7.已知函数f(x)=-cos2x-sinx+1,若$\frac{π}{4}$<α<$\frac{π}{2}$,0<β<$\frac{π}{4}$,f($\frac{π}{4}$+α)=-$\frac{4}{25}$,f($\frac{3π}{4}$+β)=-$\frac{12}{169}$,则sin(α+β)值为$\frac{8\sqrt{42}+3}{65}$.分析 由平方关系可得f(x)=-cos2x-sinx=sin2x-sinx,分别化简f($\frac{π}{4}$+α)=-$\frac{4}{25}$,f($\frac{3π}{4}$+β)=-$\frac{12}{169}$,求出sin($\frac{π}{4}$+α)和sin($\frac{3π}{4}+β$)的值,根据α、β的范围和正弦函数的性质进行取舍,由平方关系分别求出cos($\frac{π}{4}$+α)、cos($\frac{3π}{4}+β$)的值,由两角和的正弦公式求出sin[($\frac{π}{4}$+α)+($\frac{3π}{4}$+β)]的值,利用诱导公式化简得到答案.
解答 解:由题意得,f(x)=-cos2x-sinx=sin2x-sinx,
所以f($\frac{π}{4}$+α)=sin2($\frac{π}{4}$+α)-sin($\frac{π}{4}$+α)=-$\frac{4}{25}$,
则sin2($\frac{π}{4}$+α)-sin($\frac{π}{4}$+α)+$\frac{4}{25}$=0,
即[sin($\frac{π}{4}$+α)-$\frac{4}{5}$][sin($\frac{π}{4}$+α)-$\frac{1}{5}$]=0,
解得sin($\frac{π}{4}$+α)=$\frac{4}{5}$或sin($\frac{π}{4}$+α)=$\frac{1}{5}$,
因为$\frac{π}{4}$<α<$\frac{π}{2}$,所以$\frac{π}{2}<$$\frac{π}{4}$+α$<\frac{3π}{4}$,则sin($\frac{π}{4}$+α)>$\frac{\sqrt{2}}{2}$,
所以sin($\frac{π}{4}$+α)=$\frac{4}{5}$,则cos($\frac{π}{4}$+α)=-$\frac{3}{5}$,
同理f($\frac{3π}{4}$+β)=sin2($\frac{3π}{4}$+β)-sin($\frac{3π}{4}$+β)=-$\frac{12}{169}$,
则sin2($\frac{3π}{4}$+β)-sin($\frac{3π}{4}$+β)+$\frac{12}{169}$=0,
即[sin($\frac{3π}{4}$+β)-$\frac{12}{13}$][sin($\frac{3π}{4}$+β)-$\frac{1}{13}$]=0,
解得sin($\frac{3π}{4}$+β)=$\frac{12}{13}$或sin($\frac{3π}{4}$+β)=$\frac{1}{13}$,
因为0<β<$\frac{π}{4}$,所以$\frac{3π}{4}$<$\frac{3π}{4}$+β<π,则sin($\frac{3π}{4}$+β)<$\frac{\sqrt{2}}{2}$,
所以sin($\frac{3π}{4}$+β)=$\frac{1}{13}$,则cos($\frac{3π}{4}$+β)=$-\frac{2\sqrt{42}}{13}$
则sin[($\frac{π}{4}$+α)+($\frac{3π}{4}$+β)]=sin($\frac{π}{4}$+α)cos($\frac{3π}{4}$+β)+cos($\frac{π}{4}$+α)sin($\frac{3π}{4}$+β)
=$\frac{4}{5}$×($-\frac{2\sqrt{42}}{13}$)+(-$\frac{3}{5}$)×$\frac{1}{13}$=$-\frac{8\sqrt{42}+3}{65}$,
所以sin(α+β)=-sin(π+α+β)=$\frac{8\sqrt{42}+3}{65}$,
故答案为:$\frac{8\sqrt{42}+3}{65}$.
点评 本题考查两角和的正弦公式,同角三角函数的基本关系,正弦函数的性质、符号,注意三角函数的符号,考查化简、计算能力.
A. | {a|a≥2} | B. | {a|a≤2} | C. | {a|-1≤a≤2} | D. | {a|-1≤a<2} |