Question

Find the range of the following function

5 sinx + 12 cosx + 15

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Answer 1

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

Given function is

$\rm \: f(x) = 5 \: sinx \: + \: 12 \: cosx \: + \: 15$

To find the range of the function, Let's first convert the given function in to one Trigonometric function.

So, multiply and divide the given function by square root of sum of square of coefficient of sinx and square of coefficient of cosx.

That means

$\rm \: Multiply\:and\:divide\:by\: \sqrt{ {5}^{2} + {12}^{2} } = \sqrt{25 + 144} = 13$

So, above expression can be rewritten as

$\rm \: f(x) = 13\bigg(\dfrac{5}{13} \: sinx \: + \: \dfrac{12}{13} \: cosx\bigg) \: + \: 15$

Let assume that

$\rm \: cosy = \dfrac{5}{13}$

$\rm \: cos^{2} y = \dfrac{25}{169}$

$\rm \: 1 - sin^{2} y = \dfrac{25}{169}$

$\rm \: - sin^{2} y = \dfrac{25}{169} - 1$

$\rm \: - sin^{2} y = \dfrac{25 - 169}{169}$

$\rm \: - sin^{2} y = \dfrac{- 144}{169}$

$\rm \: sin^{2} y = \dfrac{144}{169}$

$\rm\implies \:siny = \dfrac{12}{13}$

So, above expression can be rewritten as

$\rm \: f(x) = 13(sinx \: cosy \: + \: siny \: cosx) + 15$

$\rm \: f(x) = 13\: sin(x + y) + 15$

We know,

$\boxed{\tt{ \: \: - 1 \leqslant sinx \leqslant 1 \: \: }}$

So, using this, we get

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

On multiply by 13 each term, we get

$\rm \: - 13 \: \leqslant \: 13sin(x + y) \: \leqslant \: 13$

On adding 15 in each term, we get

$\rm \: 15 - 13 \: \leqslant \: 13sin(x + y) + 15 \: \leqslant \: 13 + 15$

$\rm \: 2 \: \leqslant \: 13sin(x + y) + 15 \: \leqslant \: 28$

$\rm\implies \:2 \leqslant f(x) \leqslant 28$

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Short Cut trick :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf asinx + bcosx & \sf [ - \sqrt{ {a}^{2} + {b}^{2} }, \: \: \sqrt{ {a}^{2} + {b}^{2} }] \\ \\ \sf asinx + bcosx + c & \sf [c - \sqrt{ {a}^{2} + {b}^{2} }, \: c + \sqrt{ {a}^{2} + {b}^{2} }] \end{array}} \\ \end{gathered}$

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ADDITIONAL INFORMATION

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

Answer 2

${\mathfrak {\underline{\red{Given:-}}}}$

• 5 sinx + 12 cosx + 15

${\mathfrak {\underline{\red{Solution:-}}}}$

$\rm :\longmapsto\fbox{Answer in the above attachment}$

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@Shivam

#BeBrainly