Question

Find the range of the function:
f(x) = 3cos x - 4sin x

-3 ≤ 3cos x ≤ 3 and -4 ≤ -4sin x ≤ 4 Can we add these both inequalities to the range of the function?
By adding the inequalities, we get -7 ≤ 3cos x - 4sin x ≤ 7.
But the answer is given as [-5, 5]

Answer 1

Step-by-step explanation:

$f(x) = 3 \cos(x) - 4 \sin(x) \\ \\ = 5 \{ \frac{3}{5} \cos(x) - \frac{4}{5} \sin(x) \} \\ \\ let \: \cos( \alpha ) = \frac{3}{5} \: then \: \sin( \alpha ) = \frac{4}{5} \\ \\ f(x) = 5 \{ \cos( \alpha ) \cos(x) - \sin( \alpha ) \sin(x) \} \\ \\ \red {\boxed{f(x) = 5 \cos(x + \alpha ) }} \\ \\ range \: of \: cosine \: function \\ \\ - 1 \leqslant \cos(x + \alpha ) \leqslant 1 \\ \\ \implies \: - 5 \leqslant 5 \cos(x + \alpha ) \leqslant 5 \\ \\ \implies \: - 5 \leqslant f(x) \leqslant 5$

No we can't add these two inequalities to get exact range because both functions are not independent of each other . If there were cos x and sin y then you could add them and answer

[ -7 , 7 ] could have been right . [ Although it does give a interval in which our original range lies ]

Adding these two equations is meaning to say that adding maximum value of both 3cosx and -4sinx is maximum of their sum but you see they both can't be maximum at same x .

$3 \cos(x) = 1 \: \: and \: \: - 4 \sin(x) = 1$

Both sine and cosine attains their max Value at different x .

Answer 2

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

Given function is

$\rm \: f(x) = 3 cosx - 4 sinx$

To find the range of f(x) = 3 cosx - 4 sinx, we have first transform this expression to single Trigonometric function.

So, for that, we have to multiply and divide by square root of (square of coefficient of cosx + square of coefficient of sinx).

So,

$\rm \: Multiply\:and\:divide\:by \: \sqrt{ {3}^{2} + {( - 4)}^{2} } = \sqrt{9 + 16} = 5$

So, above expression can be rewritten as

$\rm \: f(x) = 5\bigg(\dfrac{3}{5} cosx - \dfrac{4}{5} sinx\bigg)$

Let assume that,

$\rm \: cosy = \dfrac{3}{5}$

So,

$\rm \: siny = \sqrt{1 - {cos}^{2} y}$

$\rm \: = \: \sqrt{1 - {\bigg(\dfrac{3}{5} \bigg) }^{2} }$

$\rm \: = \: \sqrt{1 - {\dfrac{9}{25}} }$

$\rm \: = \: \sqrt{{\dfrac{25 - 9}{25}} }$

$\rm \: = \: \sqrt{{\dfrac{16}{25}} }$

$\rm\implies \:siny = \dfrac{4}{5}$

So, above expression can be rewritten as

$\rm \: f(x) = 5(cosx \: cosy \: - \: sinx \: siny)$

$\rm \: f(x) =5 \: cos(x + y)$

Now, We know that,

$\rm \: - 1 \leqslant cos(x + y) \leqslant 1$

$\rm\implies \:\rm \: - 5 \leqslant 5cos(x + y) \leqslant 5$

$\rm\implies \:\rm \: - 5 \leqslant f(x) \leqslant 5$

Hence,

$\rm\implies \: \: \boxed{\rm{ \: - 5 \leqslant \: 3 cosx - 4 sinx \: \leqslant 5 \: \: \: }}$

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As it is true that,

-3 ≤ 3cos x ≤ 3 and -4 ≤ -4 sin x ≤ 4

But we can't add them to get [- 7, 7] as both Trigonometric functions are depend on the same variable x and both cannot assume the same values at the same value of x. For example sin0 is 0 and cos0 = 1, so these two inequalities cannot be added to give [- 7, 7]

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ADDITIONAL INFORMATION

1. The maximum and minimum value of the

$\rm \: f(x) = a \: sinx \: + \: b \: cosx \: is \: \\ \\ \rm \: [ - \sqrt{ {a}^{2} + {b}^{2}}, \: \sqrt{ {a}^{2} + {b}^{2}}]$

$\rm \: f(x) = a \: sinx \: + \: b \: cosx \: + c \: is \: \\ \\ \rm \: [c - \sqrt{ {a}^{2} + {b}^{2}}, \: c + \sqrt{ {a}^{2} + {b}^{2}}]$

2. Range of Trigonometric functions

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = sinx & \sf   - 1 \leqslant y \leqslant 1\\ \\ \sf y = cosx & \sf  - 1 \leqslant y \leqslant 1 \\ \\ \sf y = tanx & \sf y \:  \in \: ( -  \infty , \infty )\\ \\ \sf y = cosec & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = secx & \sf y \leqslant  - 1 \:  \: or \:  \: y \geqslant 1\\ \\ \sf y = cotx & \sf y \:  \in \: ( -  \infty , \infty ) \end{array}} \\ \end{gathered}$