The maximum value of 5 cos 0+3 cos(0+3)+3
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Step-by-step explanation:
(1)First convert the expression into standard form like… asinx+bcosx+c
(2)Then, min value= -√(a^2+b^2) + c
& max value= √(a^2+b^2) + c
So, 5cosx +3cos(x+60) +3
= 5cosx +3(cosxcos60 - sinxsin60) +3
= 5cosx + 3( cosx . 1/2 - sinx . √3/2 ) +3
= 5cosx + 3/2 cosx - 3 √3/2 sinx +3
= 13/2 cosx - 3 √3/2 sinx +3
Above expression is in standard form
Now, since minimum value
= - √(a^2+b^2) +c
= - √ {(13/2)^2 + (-3 √3/2)^2 } + 3
= - √ {169/4 + 27/4} +3
= - √ (196/4 ) +3
= - 14/2 +3
= -7 +3
= -4 . . . . . . . . min value
Since max value = √(a^2+b^2) + c
= 7 + 3 ( calculation , shown above)
= 10 . . . . . . . max value
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