Question

Find the extreme values of 13cos x +3√3
Sin x-4​

Answer 1

Step-by-step explanation:

$13cos x +3√3</p><p>Sin x-4$

$= (\sqrt{ {(13)}^{2} + {(3 \sqrt{3}) }^{2} } (13cos x +3√3</p><p>Sin x-4) \div (\sqrt{ {(13)}^{2} + {(3 \sqrt{3}) }^{2} })) - 4$

$= 14 \times( (13 \cos \: x \: + 3 \sqrt{3} \sin \: x) \div 14) - 4 \\ = 14 \times ((sin \: \alpha \times \cos \: x + \cos \: \alpha \times \sin \: x)$

where,

$\tan \: \alpha = 13 \div 3 \sqrt{3}$

NOW,

$= 14 \times( (13 \cos \: x \: + 3 \sqrt{3} \sin \: x) \div 14) - 4 \\ = 14 \times ((sin \: \alpha \times \cos \: x + \cos \: \alpha \times \sin \: x)$

$= 14 \times \sin( \alpha + x) - 4$

And we know that,

$- 1 \leqslant \sin(z) \leqslant 1$

this implies, above simplification is

$= 14 \times \sin( \alpha + x) - 4 \\ - 1 \leqslant \sin( \alpha + x) \leqslant 1 \\ = > - 14 \leqslant 14 \times \sin( \alpha + x) \leqslant 14 \\ = > - 14 - 4 \leqslant 14 \times \sin( \alpha + x) - 4 \leqslant 14 - 4 \\ = > - 18 \leqslant 14 \times \sin( \alpha + x) -4 \leqslant 10$

ANS.