Question

if y = (4sinx-3cosx) the the maximum value of y is​

Answer 1

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

Given function is

$\rm :\longmapsto\:y = 4sinx - 3cosx$

We know,

$\boxed{ \tt{ \: \sqrt{ {4}^{2} + {( - 3)}^{2} } = \sqrt{16 + 9} = \sqrt{25} = 5 \: }}$

On multiply and divide by 5, we get

$\rm :\longmapsto\:y = \dfrac{4}{5}sinx \: - \: \dfrac{3}{5}cosx$

Let assume that,

$\rm :\longmapsto\:\boxed{ \tt{ \: cosa = \frac{4}{5}}}$

So,

$\rm :\longmapsto\:\boxed{ \tt{ \: sina = \frac{3}{5}}}$

So, the above equation can be rewritten as

$\rm :\longmapsto\:y = 5[sinxcosa \: - \: sinacosx]$

We know,

$\boxed{ \tt{ \: sinxcosy - sinycosx = sin(x - y) \: }}$

So, using this identity, we get

$\rm :\longmapsto\:y = 5 \: sin(x - a)$

We know,

For every x belongs to real number,

$\rm :\longmapsto\: - 1 \leqslant sin(x - a) \leqslant 1$

On multiply by 5, we get

$\rm :\longmapsto\: - 5 \leqslant 5sin(x - a) \leqslant 5$

$\bf\implies \: - 5 \leqslant y \leqslant 5$

$\bf\implies \:y \: has \: maximum \: value = 5$

Note :-

Short cut

1. If y = a sinx + b cosx, then

$\rm :\longmapsto\:\boxed{ \tt{ \: -\sqrt{ {a}^{2}+{b}^{2}} \leqslant y \leqslant \sqrt{ {a}^{2}+{b}^{2}} \: }}$

2. If y = a sinx + b cosx + c, then

$\rm :\longmapsto\:\boxed{ \tt{ \: c-\sqrt{ {a}^{2}+{b}^{2}} \leqslant y \leqslant c + \sqrt{ {a}^{2}+{b}^{2}} \: }}$

Answer 2

Answer:

Maximum value of function 5