How do you find the value of the expression $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ ?$
Answer
591.6k+ views
Hint: We will use some trigonometric identities to find the value of the given expression. We will use the trigonometric identity $\sin x\cos y-\cos x\sin y=\sin \left( x-y \right).$ Then we will use the identity $\sin 300{}^\circ =\sin \left( 360-60 \right)=-\sin 60{}^\circ .$
Complete step by step solution:
Let us consider the given trigonometric expression $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ .$
We are asked to find the value of this trigonometric equation.
To find the value of this equation, we will use some familiar trigonometric identities.
We know the trigonometric identity given by $\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y.$
We can see that the given expression resembles the right-hand side of the above written identity.
When we compare the values, we will get $x=330{}^\circ $ and $y=30{}^\circ .$
So, we can equate the right-hand side of the above written trigonometric identity with the given expression by replacing the variables from the identity with the values in the expression.
So, we will get the right-hand side of the identity as $\sin \left( 330-30 \right).$
Therefore, we can equate the given expression with the above written expression to find the value of the expression.
We will get $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =\sin \left( 330-30 \right).$
We know that $330-30=300.$ So, the above obtained trigonometric equation will become $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =\sin 300{}^\circ .$
We know that $300=360-60.$
Therefore, we will get the above equation as $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =\sin \left( 360-60 \right).$
Also, we have the identity $\sin \left( 360-x \right)=-\sin x.$
So, we will get $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =-\sin 60{}^\circ .$
We know that $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}.$
Therefore, we will get $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =-\dfrac{\sqrt{3}}{2}.$
Note: There is an alternative method to find the value of the given expression. We know that $360-30=330.$ Therefore, $\sin 330{}^\circ =\sin \left( 360-30 \right)=-\sin 30{}^\circ .$ Similarly, we will get $\cos 330{}^\circ =\cos \left( 360-30 \right)=\cos 30{}^\circ .$ We know that $\sin 30{}^\circ =\dfrac{1}{2}$ and $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}.$ We will apply these values in the given expression to get $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =-\sin 30{}^\circ \cos 30{}^\circ -\cos 30{}^\circ \sin 30{}^\circ .$ And we will get $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}=-2\dfrac{\sqrt{3}}{4}=-\dfrac{\sqrt{3}}{2}.$
Complete step by step solution:
Let us consider the given trigonometric expression $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ .$
We are asked to find the value of this trigonometric equation.
To find the value of this equation, we will use some familiar trigonometric identities.
We know the trigonometric identity given by $\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y.$
We can see that the given expression resembles the right-hand side of the above written identity.
When we compare the values, we will get $x=330{}^\circ $ and $y=30{}^\circ .$
So, we can equate the right-hand side of the above written trigonometric identity with the given expression by replacing the variables from the identity with the values in the expression.
So, we will get the right-hand side of the identity as $\sin \left( 330-30 \right).$
Therefore, we can equate the given expression with the above written expression to find the value of the expression.
We will get $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =\sin \left( 330-30 \right).$
We know that $330-30=300.$ So, the above obtained trigonometric equation will become $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =\sin 300{}^\circ .$
We know that $300=360-60.$
Therefore, we will get the above equation as $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =\sin \left( 360-60 \right).$
Also, we have the identity $\sin \left( 360-x \right)=-\sin x.$
So, we will get $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =-\sin 60{}^\circ .$
We know that $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}.$
Therefore, we will get $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =-\dfrac{\sqrt{3}}{2}.$
Note: There is an alternative method to find the value of the given expression. We know that $360-30=330.$ Therefore, $\sin 330{}^\circ =\sin \left( 360-30 \right)=-\sin 30{}^\circ .$ Similarly, we will get $\cos 330{}^\circ =\cos \left( 360-30 \right)=\cos 30{}^\circ .$ We know that $\sin 30{}^\circ =\dfrac{1}{2}$ and $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}.$ We will apply these values in the given expression to get $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =-\sin 30{}^\circ \cos 30{}^\circ -\cos 30{}^\circ \sin 30{}^\circ .$ And we will get $\sin 330{}^\circ \cos 30{}^\circ -\cos 330{}^\circ \sin 30{}^\circ =-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}=-2\dfrac{\sqrt{3}}{4}=-\dfrac{\sqrt{3}}{2}.$
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

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

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

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

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

