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Find maximum and minimum value for cos^2x-6sinxcosx+3 sin^2x+2.

Leafyrito , 9 Years ago
Grade 10
anser 2 Answers
Himanshu

Last Activity: 9 Years ago

Cos2x = (1 + cos2x) / 2
Sin2x = (1 – cos2x) / 2
2sinx.cosx = sin2x
Hence, now your equation can be written as:-
=(1 + cos2x) / 2 – 3sin2x + 3(1 – cos2x) / 2 + 2
=(1/2 – 3/2)cos2x – 3sin2x +4
= – cos2x – 3sin2x +4
Now, Minimum value = – [(-1)2 + (-3)2]1/2 + 4 = 4 – 101/2
Also, Maximum value = [(-1)2 + (-3)2]1/2 + 4 = 4 + 101/2
Approved
Arnav

Last Activity: 4 Years ago

Cos2x = (1 + cos2x) / 2
 
Sin2x = (1 – cos2x) / 2
 
2sinx.cosx = sin2x
 
Hence, now your equation can be written as:-
=(1 + cos2x) / 2 – 3sin2x + 3(1 – cos2x) / 2 + 2
=(1/2 – 3/2)cos2x – 3sin2x +4
= – cos2x – 3sin2x +4
 
Now, Minimum value = – [(-1)2 + (-3)2]1/2 + 4 = 4 – 101/2
 
Also, Maximum value = [(-1)2 + (-3)2]1/2 + 4 = 4 + 101/2
 
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