Question

What will the minimum value of
$3cos(A+π/3)$



I found it -3 ..Is it correct?​

Answer 1

$\large\underline{\sf{Solution-}}$

➢ Given that

$\rm :\longmapsto\:3 \: cos\bigg(A + \dfrac{\pi}{3} \bigg)$

We know,

$\underbrace{\boxed{\sf{cos(x + y) = cosxcosy - sinxsiny}}}$

So, using this identity, we get

$\rm \: = \: \: 3\bigg(cosA \: cos\dfrac{\pi}{3} \: - \: sinA \: sin\dfrac{\pi}{3} \bigg)$

$\rm \: = \: \: 3\bigg(cosA \: \times \dfrac{1}{2} \: - \: sinA \: \times \dfrac{ \sqrt{3} }{2} \bigg)$

$\rm \: = \: \: \dfrac{3}{2}cosA - \dfrac{3 \sqrt{3} }{2}sinA$

So, it means

$\rm \implies\:\rm \:3cos\bigg(A + \dfrac{\pi}{3} \bigg) = \: \: \dfrac{3}{2}cosA - \dfrac{3 \sqrt{3} }{2}sinA$

Now, we know that

$\red{\rm :\longmapsto\: -\sqrt{ {x}^{2}+{y}^{2}} \leqslant xcost+ysint \leqslant\sqrt{ {x}^{2}+{y}^{2} } }$

So,

It implies,

$\rm :\longmapsto\:xcost \: + \: ysint \: has \: minimum \: value \: - \sqrt{ {x}^{2} + {y}^{2} }$

Hence,

Minimum value of

$\rm :\longmapsto\: \: \: \dfrac{3}{2}cosA - \dfrac{3 \sqrt{3} }{2}sinA$

$\rm \: = \: \: - \sqrt{ {\bigg(\dfrac{3}{2} \bigg) }^{2} + {\bigg(\dfrac{ - 3 \sqrt{3} }{2} \bigg) }^{2} }$

$\rm \: = \: \: - \sqrt{ {\bigg(\dfrac{9}{4} \bigg) } + {\bigg(\dfrac{27}{4} \bigg) }}$

$\rm \: = \: \: - \sqrt{ {\bigg(\dfrac{9 + 27}{4} \bigg) } }$

$\rm \: = \: \: - \sqrt{ {\bigg(\dfrac{36}{4} \bigg) } }$

$\rm \: = \: \: - \sqrt{9}$

$\rm \: = \: \: - \: 3$

Therefore,

Minimum value of

$\red{\bf :\longmapsto\:3 \: cos\bigg(A + \dfrac{\pi}{3} \bigg) = - \: 3}$

Additional Information :-

The value of f(x) = a sinx + b cosx + c always lies from

$\rm :\longmapsto\:c - \sqrt{ {a}^{2} + {b}^{2}} \leqslant f(x) \leqslant c + \sqrt{ {a}^{2} + {b}^{2} }$

The Minimum value of f(x) = a sinx + b cosx + c is

$\rm :\longmapsto\:c - \sqrt{ {a}^{2} + {b}^{2}}$

The Maximum value of f(x) = a sinx + b cosx + c is

$\rm :\longmapsto\:c + \sqrt{ {a}^{2} + {b}^{2}}$