If $sin A-cos A=0$, then $sin 2A=?$

Question

Accepted Answer

The correct option is 3:
To solve this problem, we need to use fundamental trigonometric identities and properties.

First, let's analyze the given condition:
$ \sin A - \cos A = 0 $

Step 1: Rearrange the given equation.
Add $ \cos A $ to both sides of the equation:
$ \sin A = \cos A $

Step 2: Find the value of $ 	an A $.
Divide both sides by $ \cos A $ (assuming $ \cos A 
eq 0 $). If $ \cos A = 0 $, then $ \sin A = \pm 1 $, which would mean $ \sin A 
eq \cos A $, so $ \cos A $ cannot be $ 0 $.
$ \frac{\sin A}{\cos A} = 1 $
Since $ \frac{\sin A}{\cos A} = 	an A $, we have:
$ 	an A = 1 $

Step 3: Identify the angle A for which $ 	an A = 1 $.
The general solution for $ 	an A = 1 $ is $ A = n\pi + \frac{\pi}{4} $, where $ n $ is an integer. For simplicity, we can consider the principal value $ A = \frac{\pi}{4} $ or $ A = 45^\circ $.

Step 4: Use the double angle identity for $ \sin 2A $.
We need to find the value of $ \sin 2A $. The double angle identity for sine is:
$ \sin 2A = 2 \sin A \cos A $

Step 5: Substitute $ \sin A = \cos A $ into the double angle identity.
Since we found that $ \sin A = \cos A $, we can substitute $ \cos A $ with $ \sin A $ (or vice versa):
$ \sin 2A = 2 \sin A (\sin A) $
$ \sin 2A = 2 \sin^2 A $

Alternatively, substitute $ \sin A = \cos A $ to get:
$ \sin 2A = 2 \cos A (\cos A) $
$ \sin 2A = 2 \cos^2 A $

Let's work from $ \sin A = \cos A $.
We know that $ \sin^2 A + \cos^2 A = 1 $.
Substitute $ \cos A = \sin A $:
$ \sin^2 A + \sin^2 A = 1 $
$ 2 \sin^2 A = 1 $
$ \sin^2 A = \frac{1}{2} $
So, $ \sin A = \pm \frac{1}{\sqrt{2}} $.
Since $ \sin A = \cos A $, both $ \sin A $ and $ \cos A $ must have the same sign. This means $ A $ is in the first or third quadrant.

Using $ \sin 2A = 2 \sin A \cos A $:
Since $ \sin A = \cos A $, we can substitute $ \cos A $ with $ \sin A $:
$ \sin 2A = 2 \sin A \cdot \sin A = 2 \sin^2 A $
From $ 2 \sin^2 A = 1 $, we directly get $ \sin 2A = 1 $.

Let's also verify this using the value of A from Step 3:
If $ A = 45^\circ $, then:
$ \sin A = \sin 45^\circ = \frac{1}{\sqrt{2}} $
$ \cos A = \cos 45^\circ = \frac{1}{\sqrt{2}} $
So, $ \sin A - \cos A = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 $, which matches the given condition.

Now, calculate $ \sin 2A $ for $ A = 45^\circ $:
$ \sin 2A = \sin (2 	imes 45^\circ) = \sin 90^\circ $
We know that $ \sin 90^\circ = 1 $.

Therefore, $ \sin 2A = 1 $.

The incorrect option is 0:
This would imply $ \sin A = 0 $ or $ \cos A = 0 $ which is not consistent with the given condition $ \sin A - \cos A = 0 $ (unless $ A = 0 $ and $ \cos A = 1 $ or $ A = \frac{\pi}{2} $ and $ \sin A = 1 $, neither of which makes $ \sin A - \cos A = 0 $). Also, $ \sin 2A = 0 $ would imply $ 2A = n\pi $. If $ A = \frac{\pi}{4} $, then $ 2A = \frac{\pi}{2} $, and $ \sin(\frac{\pi}{2}) 
eq 0 $.

The incorrect option is -1:
This would occur if $ 2A = \frac{3\pi}{2} + 2n\pi $. If $ A = \frac{\pi}{4} $, $ 2A = \frac{\pi}{2} $, and $ \sin(\frac{\pi}{2}) 
eq -1 $.

The incorrect option is $\frac{1}{2}$:
This value is incorrect because $ \sin 2A = 1 $ as derived from the given condition. This might be a result of misremembering trigonometric values or identities. For example, if one mistakenly thought $ A=30^\circ $, then $ \sin A - \cos A 
eq 0 $, but $ \sin 2A = \sin 60^\circ = \frac{\sqrt{3}}{2} $. If one might have confused $ \sin^2 A $ with $ \sin 2A $, as $ \sin^2 A = \frac{1}{2} $.