Question

Find the minimum value of 5cosA + 12sinA + 12​

Answer 1

Given :

5 cos A + 12 sin A + 12

To Find :

Minimum value of the given

Solution :

Minimum value of ' a cos A + b sin B ' is - √(a² + b²) as well ,

Maximum value of ' a cos A + b sin B '  is + √(a² + b²)

Lets calculate the minimum value of 5 cos A + 12 sin A

➠ - √(5² + 12²)

➠ - √(25 + 144)

➠ - √(169)

➠ - 13

Minimum value of 12 is nothing but 12

Minimum value of 5 cos A + 12 sin A + 12 is ,

➙ 5 cos A + 12 sin A + 12

➙ - 13 + 12

➙ - 1   $\pink{\bigstar}$

Alternate Method :

$:\to \sf 5\ cos\ A+12\ sin\ A+12$

A/c to Pythagoras triplet 3rd number will be 13 ,

Divide and multiply with 13 ,

$:\to \sf \dfrac{13}{13}(5\ cos\ A+12\ sin\ A)+12$

$:\to \sf 13\bigg(\dfrac{5}{13}.cos\ A+\dfrac{12}{13}.sin\ A\bigg)+12$

Let , cos B = 5/13 and sin B = 12/13

$:\to \sf 13(cos\ B.cos\ A+sin\ B.sin\ A)+12$

$:\to \sf 13(cos\ A.cos\ B+sin\ A.sin\ B)+12$

$:\to \sf 13(cos(A-B))+12$

We know that , ' Minimum value for cos θ is - 1 as well maximum value of + 1 '

$:\to \sf 13(-1)+12$

$:\to \sf -13+12$

$:\to \sf -1\ \; \green{\bigstar}$

Answer 2

See, the minimum value of sin(x) and cos(x) is -1 for both ( and ofcourse the maximum value is +1)

So for the expression to take minimum value, sin(x) and cos(x) must take their minimum value i.e. -1.

So substituting that in the eqn we get, -5 -12 +12

Which gives the answer -5.