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Question: How do you use the angle sum identity to find the exact value of \[\sin ({195^0})\] ?...

How do you use the angle sum identity to find the exact value of sin⁡(1950)\sin ({195^0}) ?

Explanation

Solution

Hint : Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions. The angle sum identity of sine function is sin⁡(a+b)=sin⁡(a).cos⁡(b)+cos⁡(a).sin⁡(b)\sin (a + b) = \sin (a).\cos (b) + \cos (a).\sin (b) . In the given problem we have angel value 195. We can express 195 as a sum of 45 and 150. Using the formula we can solve this.

Complete step-by-step answer :
Given.
sin⁡(1950)\sin ({195^0})
We need 195 as the sum of two numbers.
If we express 195 as the sum of 15 and 180. Then we need to know the value of sin⁡(150)\sin ({15^0}) .
If we express 195 as the sum of 30 and 165. Then we need to know the value of sin⁡(1650)\sin ({165^0}) and cos⁡(1650)\cos ({165^0}) . But we don’t know the values of those.
If we express 195 as the sum of 45 and 150. Then we need to know the value of sin⁡(1500)\sin ({150^0}) and cos⁡(1500)\cos ({150^0}) .
sin⁡(1500)=sin⁡(1800−300)\sin ({150^0}) = \sin ({180^0} - {30^0})
(Supplementary angle)
=sin⁡(300)= \sin ({30^0})
sin⁡(1500)=12\sin ({150^0}) = \dfrac{1}{2} .
Similarly
cos⁡(1500)=cos⁡(1800−300)\cos ({150^0}) = \cos ({180^0} - {30^0})
=−cos⁡(300)= - \cos ({30^0})
Negative value is because cosine is negative in the second quadrant.
cos⁡(1500)=−32\cos ({150^0}) = - \dfrac{{\sqrt 3 }}{2}
Now we have sin⁡(195)\sin (195) .
We can express 1950=450+1500{195^0} = {45^0} + {150^0}
=sin⁡(450+1500)= \sin ({45^0} + {150^0})
Using the identity sin⁡(a+b)=sin⁡(a).cos⁡(b)+cos⁡(a).sin⁡(b)\sin (a + b) = \sin (a).\cos (b) + \cos (a).\sin (b) . We have,
=sin⁡(450)cos⁡(1500)+cos⁡(450)sin⁡(1500)= \sin ({45^0})\cos ({150^0}) + \cos ({45^0})\sin ({150^0})
Substituting the known values we have,
=(12)(−32)+(12)(12)= \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( { - \dfrac{{\sqrt 3 }}{2}} \right) + \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{1}{2}} \right)
=−322+122= \dfrac{{ - \sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }}
Taking LCM we have,
=−3+122= \dfrac{{ - \sqrt 3 + 1}}{{2\sqrt 2 }}
Hence, ⇒sin⁡(1950)=−3+122 \Rightarrow \sin ({195^0}) = \dfrac{{ - \sqrt 3 + 1}}{{2\sqrt 2 }}
So, the correct answer is “−3+122\dfrac{{ - \sqrt 3 + 1}}{{2\sqrt 2 }}”.

Note : Remember the formula sine and cosine addition and subtract formula well. In case id sine if we have cosine we will use the formula cos⁡(a+b)=cos⁡(a).cos⁡(b)−sin⁡(a).sin⁡(b)\cos (a + b) = \cos (a).\cos (b) - \sin (a).\sin (b) . Similarly we have sin⁡(a−b)=sin⁡(a).cos⁡(b)−cos⁡(a).sin⁡(b)\sin (a - b) = \sin (a).\cos (b) - \cos (a).\sin (b) and cos⁡(a−b)=cos⁡(a).cos⁡(b)+sin⁡(a).sin⁡(b)\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b) . Depending on the angle we use the required formulas. Also remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.