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# Find the maximum and min (wherever possible) value of: 1) cos^2 x - 6 sin x cos x + 3sin^2 X + 2 2) cos 2x + cos x 3) cos^2(pi/4 + x)+ (sinx - cosx)^2

147 Points
11 years ago

Dear Vishant

cos^2 x - 6 sin x cos x + 3sin^2 X + 2

=(1+cos2x)/2  -3sin2x +3(1-cos2x)/2 +2

= -cos2x -3sin2x +4

we know    -√a2+b2 ≤ asinx + bcosx ≤√a2+b2

so      -√10 ≤   -cos2x -3sin2x√10

so   -√10  +4  ≤   -cos2x -3sin2x  +4 ≤√10 +4

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