prove that cos[(3π/4)+x]-cos[(3π/4)-3]=-√2sinx
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=cos{3π/4+x}-cos{3π/4-x}
=cos(3π/4)cosx-sin(3π/4)sinx-[cos(3π/4)cosx+sin(3π/4)sinx}
=-cosx/√2-sinx/√2+cosx/√2-sinx/√2
=-2sinx/√2
=-√2sinx
=cos(3π/4)cosx-sin(3π/4)sinx-[cos(3π/4)cosx+sin(3π/4)sinx}
=-cosx/√2-sinx/√2+cosx/√2-sinx/√2
=-2sinx/√2
=-√2sinx
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