Question

The value of sin ( \[{45}^° + \theta) - \cos ( {45}^°- \theta)\] is equal to 

Options

• 2 cos \[\theta\]

• 0  

•   2 sin \[\theta\]

• 1

Answer 1

We know that, 

\[\sin\left( 90 - \theta \right) = \cos\theta\]

So, 

\[\sin\left( 45°+ \theta \right) = \cos\left[ 90 - \left( 45° + \theta \right) \right] = \cos\left( 45° - \theta \right)\] 

\[\therefore \sin\left( 45°+ \theta \right) - \cos\left( 45°- \theta \right)\]
\[ = \cos\left( 45° - \theta \right) - \cos\left( 45° - \theta \right)\]
\[ = 0\]