How do you find the exact values of $ \sin {165^0} $ using the half angle formulae ?
Answer
592.8k+ views
Hint: First we will evaluate the right-hand of the equation and then further the left-hand side of the equation. We will use the following formula to evaluate $ \sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} $ and then we will further this expression to $ \sin x $ form and hence evaluate the value of the term.
Complete step-by-step answer:
We will start off by using the formula $ \sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} $ .
Hence, the equation will become,
\[
= \sin 165 \\
\sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} \\
\sin \left( {\dfrac{{330}}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos 330}}{2}} \\
\sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \cos (360 - 330)}}{2}} \\
\sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \cos (30)}}{2}} \;
\]
Now we know that the value of $ \cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2} $
So now here, the value will be,
\[
\sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \cos (30)}}{2}} \\
\sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \dfrac{{\sqrt 3 }}{2}}}{2}} \\
\sin \left( {165} \right) = \pm \sqrt {\dfrac{{2 - \sqrt 3 }}{4}} \\
\sin \left( {165} \right) = \pm \dfrac{{\sqrt {2 - \sqrt 3 } }}{2} \;
\]
So, the correct answer is “ $ \pm \dfrac{{\sqrt {2 - \sqrt 3 } }}{2} $ ”.
Note: While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the trigonometric identities $ {\sin ^2}x + {\cos ^2}x = 1 $ and $ \cos 2x = 2{\cos ^2}x - 1 $ . While opening the brackets make sure you are opening the brackets properly with their respective signs. Also remember that $ \tan x = \,\dfrac{{\sin x}}{{\cos x}} $ .
While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities.
Complete step-by-step answer:
We will start off by using the formula $ \sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} $ .
Hence, the equation will become,
\[
= \sin 165 \\
\sin \left( {\dfrac{x}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} \\
\sin \left( {\dfrac{{330}}{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos 330}}{2}} \\
\sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \cos (360 - 330)}}{2}} \\
\sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \cos (30)}}{2}} \;
\]
Now we know that the value of $ \cos \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{2} $
So now here, the value will be,
\[
\sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \cos (30)}}{2}} \\
\sin \left( {165} \right) = \pm \sqrt {\dfrac{{1 - \dfrac{{\sqrt 3 }}{2}}}{2}} \\
\sin \left( {165} \right) = \pm \sqrt {\dfrac{{2 - \sqrt 3 }}{4}} \\
\sin \left( {165} \right) = \pm \dfrac{{\sqrt {2 - \sqrt 3 } }}{2} \;
\]
So, the correct answer is “ $ \pm \dfrac{{\sqrt {2 - \sqrt 3 } }}{2} $ ”.
Note: While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the trigonometric identities $ {\sin ^2}x + {\cos ^2}x = 1 $ and $ \cos 2x = 2{\cos ^2}x - 1 $ . While opening the brackets make sure you are opening the brackets properly with their respective signs. Also remember that $ \tan x = \,\dfrac{{\sin x}}{{\cos x}} $ .
While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Find the value of the expression given below sin 30circ class 11 maths CBSE

Two of the body parts which do not appear in MRI are class 11 biology CBSE

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

