Question

solve the following 3cos thetha+8cos thetha+5cos thetha=​

Answer 1

For finding min & max values of trigonometric expression …

(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