Question

How to evaluate \[\sin \left( { - 135} \right)\]?

Answer 1

Hint: The trigonometry identity \[\sin \left( { - x} \right) = - \sin \left( x \right)\]. Here x is \[ - {135^ \circ }\] and \[ - {135^ \circ }\]is equal to 180 degree minus 45 degree. This can be substituted in place of x. The trigonometry identity \[\sin \left( {180 - x} \right) = \sin \left( x \right)\]. Here x is 45 degrees. The value of \[\sin {\left( {45} \right)^ \circ } = \dfrac{1}{{\sqrt 2 }}\]. These reduction formulae can be used to evaluate such questions. Also quadrant can be determined for confirmation of answers.
Taylor series expansion can be used each time whenever needed to compute trigonometric functions for a given angle.

Complete step by step answer:
We know that, \[\sin \left( { - x} \right) = - \sin \left( x \right)\]
So,
 \[
  \sin \left( { - 135} \right) = - \sin \left( {135} \right) \\
   \Rightarrow \sin \left( { - 135} \right) = - \sin \left( {180 - 45} \right) \\
\]
We have, \[\sin \left( {180 - x} \right) = \sin \left( x \right)\]
So, \[\sin \left( { - {{135}^ \circ }} \right) = - \sin \left( {{{45}^ \circ }} \right)\]
We know that, \[\sin \left( {{{45}^ \circ }} \right) = \dfrac{1}{{\sqrt 2 }}\]
So, \[\sin \left( { - {{135}^ \circ }} \right) = - \dfrac{1}{{\sqrt 2 }}\]

Note: Before calculation quadrant must be determined. If our angle is greater than 90 and less than or equal to 180 degrees, it is located in quadrant II. In the second quadrant the values for sin are positive only. If the angle is greater than 180 and less or equal to 270 degrees then it falls in the third quadrant where the sin functions are negative. Therefore, before solving such questions quadrants can easily be determined.

We have \[ - {135^ \circ }\] which is in quadrant 3 (going clockwise rather than going anticlockwise). The reference angle is 45 degrees where the terminal arm is 45 degrees from the X axis.

Determining the quadrant makes it easier to verify the answer whether correct or wrong.
The sine of the angle is equal to the product of length of the opposite side and length of the hypotenuse. Taylor series can be used also as an alternative way to solve any angle of a trigonometric function.