How do you solve $sin A - cos A = 0$ $sinA-cosA=0$ ?

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Answer 1 · 2015-08-04T04:19:06

Solve (sin x - cos x) = 0.

Ans: $\frac{π}{4} and \frac{5 π}{4}$ $pi/4 and (5pi)/4$

Use trig identity:
$sin x - cos x = \sqrt{2} sin (x - \frac{π}{4})$ $sin x - cos x = sqrt2sin (x - pi/4)$

$sin (x - \frac{π}{4}) = 0$ $sin (x - pi/4) = 0$

a. $(x - \frac{π}{4}) = 0$ $(x - pi/4) = 0$ --> $x = \frac{π}{4}$ $x = pi/4$

b. $(x - \frac{π}{4}) = π$ $(x - pi/4) = pi$ --> $x = π + \frac{π}{4} = \frac{5 π}{4}$ $x = pi + pi/4 = (5pi)/4$

How do you solve sinA-cosA=0sinA−cosA=0sinA-cosA=0?