How do you solve $sin 4 β = cos 5 β$ $sin 4beta = cos 5beta$ ?

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How do you solve $sin 4 β = cos 5 β$ $sin 4beta = cos 5beta$ ?

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Answer 1 · 2016-08-18T04:27:29

$\frac{π}{18} + 2 k π$ $pi/18 + 2kpi$
$\frac{3 π}{2} + 2 k π$ $(3pi)/2 + 2kpi$

Explanation:

sin 4x = cos 5x (1)
Use trig identity:
$sin a = cos (\frac{π}{2} - a)$ $sin a = cos (pi/2 - a)$
$sin 4 x = cos (\frac{π}{2} - 4 x)$ $sin 4x = cos (pi/2 - 4x)$
From (1):
$cos 5 x = cos (\frac{π}{2} - 4 x)$ $cos 5x = cos (pi/2 - 4x)$
$5 x = \pm (\frac{π}{2} - 4 x)$ $5x = +- (pi/2 - 4x)$
a. $5 x = \frac{π}{2} - 4 x$ $5x = pi/2 - 4x$ --> $9 x = \frac{π}{2}$ $9x = pi/2$ --> $x = \frac{π}{18}$ $x = pi/18$
b. $5 x = - \frac{π}{2} + 4 x$ $5x = - pi/2 + 4x$ --> $x = - \frac{π}{2}$ $x = - pi/2$ , or $x = \frac{3 π}{2}$ $x = (3pi)/2$
General answer
$x = \frac{π}{18} + 2 k π$ $x = pi/18 + 2kpi$
$x = \frac{3 π}{2} + 2 k π$ $x = (3pi)/2 + 2kpi$
Check
$x = \frac{π}{18} = 10^{\circ}$ $x = pi/18 = 10^@$ --> sin 4x = sin 40 = 0.642
cos 5x = cos 50 = 0.642. OK

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How do you solve $sin 4 β = cos 5 β$ $sin 4beta = cos 5beta$ ?

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