Math, asked by gavy5393, 8 months ago

The maximum value of 5 cos 0+3 cos(0+3)+3​

Answers

Answered by manasvibhawke
1

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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