若8cos(π/4+α)cos(π/4-α)=1,则(sinα)^4+(cosα)^4=?
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注意:这两个角的关系是互余的.即: (π/4+α)+(π/4-α)=π/2∴1=8cos(π/4+α)cos(π/4-α)=8cos[π/2-(π/4-α)]cos(π/4-α)=8sin(π/4-α)]cos(π/4-α)=4sin(π/2-2α)=4cos2α=4[(cosα)^-(sinα)^]即:(cosα)^-(sinα)^=1/4解:∵8cos(π/4+α)cos(π/4-α)=1,∴(cosα)^-(sinα)^=1/4而2[(sinα)^4+(cosα)^4]=[(sinα)^+(cosα)^]^+[(sinα)^-(cosα)^]^=1+1/16=17/16∴(sinα)^4+(cosα)^4=17/32
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