Question

How do you determine $ \sin 345 $ ?

Answer 1

Hint: We know that we cannot directly write the value for the given function. Therefore, we will convert the angle here as the addition of two other angles, the value of which is known to us. After that we will apply the formula for the sine function of addition of two angles to get our final answer.
Formula used:
 $ \sin (A + B) = \sin A\cos B + \cos A\sin B $

Complete step by step solution:
We are given the function $ \sin 345 $ .
Our first step here is to convert the angle of 345 degree into two angles whose values are known or easily obtained. Therefore, we will take 345 as the sum of 300 and 45.
Therefore our function becomes:
 $ \sin 345 = \sin \left( {300 + 45} \right) $
If we apply the formula $ \sin (A + B) = \sin A\cos B + \cos A\sin B $ , we will get,
 $ \sin (300 + 45) = \sin 300\cos 45 + \cos 300\sin 45 $
To solve this, we will now determine the values of $ \sin 300 $ , $ \cos 300 $ , $ \sin 45 $ and $ \cos 45 $ .
We know that $ \sin 300 = - \sin 60 = - \dfrac{{\sqrt 3 }}{2} $ and $ \cos 300 = \cos 60 = \dfrac{1}{2} $ .
Also, $ \sin 45 = \cos 45 = \dfrac{1}{{\sqrt 2 }} $ .
For the ease of calculation we will convert this denominator as 2 by multiplying the ratio with $ \dfrac{{\sqrt 2 }}{{\sqrt 2 }} $ .
Therefore, we get $ \sin 45 = \cos 45 = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2} $
Now, we will put all these values in the equation.
 $
  \sin (300 + 45) = \sin 300\cos 45 + \cos 300\sin 45 \\
   \Rightarrow \sin (300 + 45) = \left( { - \dfrac{{\sqrt 3 }}{2}} \right)\left( {\dfrac{{\sqrt 2 }}{2}} \right) + \left( {\dfrac{1}{2}} \right)\left( {\dfrac{{\sqrt 2 }}{2}} \right) \\
   \Rightarrow \sin 345 = - \dfrac{{\sqrt 6 }}{4} + \dfrac{{\sqrt 2 }}{4} = \dfrac{1}{4}\left( {\sqrt 2 - \sqrt 6 } \right) \;
  $
Hence, the value of $ \sin 345 $ is $ \dfrac{1}{4}\left( {\sqrt 2 - \sqrt 6 } \right) $ .
So, the correct answer is “ $ \dfrac{1}{4}\left( {\sqrt 2 - \sqrt 6 } \right) $ ”.

Note: While solving this type of question, we need to be very careful with the sign of sine and cosine function. Here, we have taken $ \sin 300 = - \sin 60 = - \dfrac{{\sqrt 3 }}{2} $ , because 300 comes in the fourth quadrant where the sine function is negative. Whereas the cosine function is positive in the fourth quadrant and therefore we have taken $ \cos 300 = \cos 60 = \dfrac{1}{2} $ .
In short we have to keep in mind the four facts:
In the first quadrant, all the trigonometric functions are positive.
In the second quadrant, only sine function is positive.
In the third quadrant, only tangent function is positive.
In the fourth quadrant, only cosine function is positive.