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Question Number 142794 by EDWIN88 last updated on 05/Jun/21

If 2 \cos (\frac{5 π}{4} + 3 x) \cos (\frac{π}{4} + 3 x) = 0 and

\sin (2 x - 2 y) = \cos y where \frac{π}{4} ⩽ x ⩽ \frac{π}{2} and

\frac{π}{4} ⩽ y ⩽ \frac{π}{2} . find the value of {\begin{cases} \sin (2 x + y) \\ \cos (2 x + y) \\ \cos (2 x - y) \\ \sin (2 x - y) \end{cases}

Answered by Rasheed.Sindhi last updated on 05/Jun/21

2 \cos (\frac{5 π}{4} + 3 x) \cos (\frac{π}{4} + 3 x) = 0

\cos (\frac{5 π}{4} + 3 x) = 0 ∣ \cos (\frac{π}{4} + 3 x) = 0

\frac{5 π}{4} + 3 x = \frac{π}{2} (2 n - 1) ∣ \frac{π}{4} + 3 x = \frac{π}{2} (2 n - 1)

n \in Z

3 x = \frac{π}{2} (2 n - 1) - \frac{5 π}{4} ∣ 3 x = \frac{π}{2} (2 n - 1) - \frac{π}{4}

x = \frac{π}{6} (2 n - 1) - \frac{5 π}{12} ∣ x = \frac{π}{6} (2 n - 1) - \frac{π}{12}

x = \frac{{2 (2 n - 1) - 5} π}{12} ∣ x = \frac{{2 (2 n - 1) - 1} π}{12}

x = \frac{(4 n - 7) π}{12} ∣ x = \frac{(4 n - 3) π}{12}

\frac{π}{4} ⩽ x ⩽ \frac{π}{2}

\frac{π}{4} ⩽ \frac{(4 n - 7) π}{12} ⩽ \frac{π}{2} ∣ \frac{π}{4} ⩽ \frac{(4 n - 3) π}{12} ⩽ \frac{π}{2}

3 π ⩽ (4 n - 7) π ⩽ 6 π ∣ 3 π ⩽ (4 n - 3) π ⩽ 6 π

3 ⩽ (4 n - 7) ⩽ 6 ∣ 3 ⩽ (4 n - 3) ⩽ 6

10 ⩽ 4 n ⩽ 13 ∣ 6 ⩽ 4 n ⩽ 9

\frac{5}{2} ⩽ n ⩽ \frac{13}{4} ∣ \frac{3}{2} ⩽ n ⩽ \frac{9}{4}

2 \frac{1}{2} ⩽ n ⩽ 3 \frac{1}{4} ∣ 1 \frac{1}{2} ⩽ n ⩽ 2 \frac{1}{4}

n = 3 ∣ n = 2

x = \frac{(4 n - 7) π}{12} ∣ x = \frac{(4 n - 3) π}{12}

x = \frac{(4 (3) - 7) π}{12} ∣ x = \frac{(4 (2) - 3) π}{12}

x = \frac{5 π}{12}

\sin (2 x - 2 y) = \cos y

\sin (2 (\frac{5 π}{12}) - 2 y) = \cos y

\sin (\frac{5 π}{6} - 2 y) = \sin (\frac{π}{2} - y)

\frac{5 π}{6} - 2 y = \frac{π}{2} - y

y = \frac{5 π}{6} - \frac{π}{2} = \frac{5 π - 3 π}{6} \Rightarrow y = \frac{π}{3}

Answered by MJS_new last updated on 05/Jun/21

2 \cos (\frac{5 π}{4} + 3 x) \cos (\frac{π}{4} + 3 x) = 0

\sin 6 x - 1 = 0 \Rightarrow x = \frac{5 π}{12}

\Rightarrow y = \frac{π}{3}

Answered by Rasheed.Sindhi last updated on 06/Jun/21

⟨_{∙}^{∙} ⟩ A n O ther W ay ⟨_{∙}^{∙} ⟩

2 \cos \underset{n o t i c e t h a t t h e d i f f i s π}{\underset{⏟}{(\frac{5 π}{4} + 3 x) \cos (\frac{π}{4} + 3 x)}} = 0

\cos (\frac{π}{4} + π + 3 x) \cos (\frac{π}{4} + 3 x) = 0

\cos (π + θ) = - \cos θ

- \cos (\frac{π}{4} + 3 x) \cos (\frac{π}{4} + 3 x) = 0

\cos^{2} (\frac{π}{4} + 3 x) = 0

\cos (\frac{π}{4} + 3 x) = 0

\cos (\frac{π}{4} + 3 x) = \cos (\frac{π}{2} (2 n - 1))

\frac{π}{4} + 3 x = \frac{π}{2} (2 n - 1)

x = \frac{1}{3} (\frac{(2 n - 1) π}{2} - \frac{π}{4}) = \frac{2 (2 n - 1) π - π}{12}

= \frac{(4 n - 3) π}{12}

\frac{π}{4} ⩽ x ⩽ \frac{π}{2}

\frac{π}{4} ⩽ \frac{(4 n - 3) π}{12} ⩽ \frac{π}{2}

3 ⩽ 4 n - 3 ⩽ 6

6 ⩽ 4 n ⩽ 9

\frac{3}{2} ⩽ n ⩽ \frac{9}{4}

1 \frac{1}{2} ⩽ n ⩽ 2 \frac{1}{4}

n = 2

x = \frac{(4 n - 3) π}{12} = \frac{{4 (2) - 3} π}{12} = \frac{5 π}{12}

\sin (2 x - 2 y) = \cos y

\sin (2 x - 2 y) = \sin (\frac{π}{2} - y)

2 x - 2 y = \frac{π}{2} - y

y = \frac{π}{3}