if-sin-x-3-cos-x-1-2-3-sin-4x-cos-4x-

Question Number 191344 by mnjuly1970 last updated on 23/Apr/23

if s i n (x) + \sqrt{3} c o s (x) = \frac{1}{2}

\Rightarrow \sqrt{3} s i n (4 x) - c o s (4 x) = ?

Answered by mehdee42 last updated on 23/Apr/23

\frac{1}{2} s i n x + \frac{\sqrt{3}}{2} c o s x = \frac{1}{4} \Rightarrow c o s (x - 30) = \frac{1}{4}

\Rightarrow c o s (2 x - 60) = 2 (\frac{1}{16}) - 1 = - \frac{7}{8}

\Rightarrow c o s (4 x - 120) = 2 (\frac{49}{16}) - 1 = \frac{41}{8} ∴ 2 (\frac{49}{64}) - 1 = \frac{17}{32}

\Rightarrow c o s 4 x c o s 120 + s i n 4 x s i n 120 = \frac{41}{8} ∴ = \frac{17}{32}

- \frac{1}{2} c o s 4 x + \frac{\sqrt{3}}{2} s i n 4 x = \frac{41}{8} ∴ = \frac{17}{32}

\Rightarrow \sqrt{3} s i n 4 x - c o s 4 x = \frac{41}{4} ∴ \frac{17}{16}

Commented by mr W last updated on 23/Apr/23

v e r y n i c e s i r!

Commented by mehdee42 last updated on 23/Apr/23

h e l l o s i r W

m y m i s t a k e w a s i n t h e c a l c u l a t i o n, w h i c h i h a v e c o r r e c t e d

w i t h r e d c o l o r . t h a n k y o u

Answered by mr W last updated on 23/Apr/23

\frac{\sin x}{2} + \frac{\sqrt{3}}{2} \cos x = \frac{1}{4}

\sin x \sin \frac{π}{6} + \cos x \cos \frac{π}{6} = \frac{1}{4}

\cos (x - \frac{π}{6}) = \frac{1}{4}

\Rightarrow x - \frac{π}{6} = 2 k π \pm \cos^{- 1} \frac{1}{4}

\Rightarrow x = 2 k π + \frac{π}{6} \pm \cos^{- 1} \frac{1}{4}

\sqrt{3} \sin 4 x - \cos 4 x

= 2 (\sin 4 x \cos \frac{π}{6} - \cos 4 x \sin \frac{π}{6})

= 2 \sin (4 x - \frac{π}{6})

= 2 \sin (8 k π + \frac{π}{2} \pm 4 \cos^{- 1} \frac{1}{4})

= 2 \cos (4 \cos^{- 1} \frac{1}{4})

= 2 [2 \cos^{2} (2 \cos^{- 1} \frac{1}{4}) - 1]

= 2 [2 {2 \cos^{2} (\cos^{- 1} \frac{1}{4}) - 1}^{2} - 1]

= 2 [2 {2 \times \frac{1}{4^{2}} - 1}^{2} - 1]

= 2 [2 {\frac{7}{8}}^{2} - 1]

= \frac{17}{16} ✓

Commented by mnjuly1970 last updated on 23/Apr/23

v e r y n i c e s o l u t i o n

Answered by cortano12 last updated on 23/Apr/23

\Rightarrow \sin^{2} x + 3 \cos^{2} x + \sqrt{3} \sin 2 x = \frac{1}{4}

\Rightarrow (\frac{1 - \cos 2 x}{2}) + 3 (\frac{1 + \cos 2 x}{2}) + \sqrt{3} \sin 2 x = \frac{1}{4}

\Rightarrow 4 + 2 \cos 2 x + 2 \sqrt{3} \sin 2 x = \frac{1}{2}

\Rightarrow 2 \cos 2 x + 2 \sqrt{3} \sin 2 x = - \frac{7}{2}

\Rightarrow 4 \cos^{2} 2 x + 12 \sin^{2} 2 x + 4 \sqrt{3} \sin 4 x = \frac{49}{4}

\Rightarrow 4 (\frac{1 + \cos 4 x}{2}) + 12 (\frac{1 - \cos 4 x}{2}) + 4 \sqrt{3} \sin 4 x = \frac{49}{4}

\Rightarrow 8 - 4 \cos 4 x + 4 \sqrt{3} \sin 4 x = \frac{49}{4}

\Rightarrow \sqrt{3} \sin 4 x - \cos 4 x = \frac{49}{16} - 2

\Rightarrow \sqrt{3} \sin 4 x - \cos 4 x = \frac{17}{16}

Commented by mnjuly1970 last updated on 23/Apr/23

t h a n k s a l o t s i r c o r t a n o ⋚

t h a n k s a l o t s i r c o r t a n o \cancel ⋚