InterviewSolution
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InterviewSolution
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Prove that: `sin^2(pi/8+A/2)-sin^2(pi/8-A/2)=1/sqrt(2)sinA` |
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Answer» `L.H.S. = sin^2(pi/8+A/2) - sin^2(pi/8-A/2)` `=(sin(pi/8)cos(A/2)+cos(pi/8)sin(A/2))^2-(sin(pi/8)cos(A/2)-cos(pi/8)sin(A/2))^2` As `a^2-b^2 = (a+b)(a-b)`, so it becomes `=(sin(pi/8)cos(A/2)+cos(pi/8)sin(A/2)+sin(pi/8)cos(A/2)-cos(pi/8)sin(A/2))(sin(pi/8)cos(A/2)-cos(pi/8)sin(A/2)-sin(pi/8)cos(A/2)+cos(pi/8)sin(A/2))` `=4(sin(pi/8)cos(A/2)(cos(pi/8)sin(A/2))` `=(2sin(pi/8)cos(pi/8))(2sin(A/2)cos(A/2))` `=sin(2*pi/8)sin(2*A/2)` `=sinpi/4sinA` `=1/sqrt2sinA = R.H.S.` |
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