
Prove the trigonometric function:
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin (x)$
Answer
612.6k+ views
Hint:In this question, we have to find the sum of cosines of two different angles. However, the angles in each term are a sum of two angles. Therefore, we should understand the formula for finding the cosine of the sum of two angles and then use it to solve this question.
Complete step-by-step answer:
Here, we are given the cosines of two different angles and we have to find their sum. However, the angle in each cosine term is itself a sum or difference of two angles. Therefore, we should first understand the equation for finding the cosine of sum and difference of two angles.
We know that the expansion of cosine of sum and difference of two angles is given by
$\begin{align}
& \cos (a+b)=\cos (a)\cos (b)-\sin (a)\sin (b)........(1.1) \\
& \cos (a-b)=\cos (a)\cos (b)+\sin (a)\sin (b)............(1.2) \\
\end{align}$
If we subtract these two equations, we obtain
$\begin{align}
& \cos (a+b)-\cos (a-b)=\cos (a)\cos (b)-\sin (a)\sin (b)-\left( \cos (a)\cos (b)+\sin (a)\sin (b) \right) \\
& \Rightarrow \cos (a+b)-\cos (a-b)=-2\sin (a)\sin (b)....................(1.3) \\
\end{align}$
We find that this equation is similar to the expression in the question. Taking \[a=\dfrac{3\pi }{4}\]and \[b=x\] in equation (1.3), we obtain
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\sin \left( \dfrac{3\pi }{4} \right)\sin (x).......(1.4)$
Now, using the trigonometric equation
$\sin \left( \dfrac{\pi }{2}+x \right)=\sin (x)$
We obtain
\[\sin \left( \dfrac{3\pi }{4} \right)=\sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}.........(1.5)\]
Thus, using this value in equation (1.4), we obtain
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\times \dfrac{1}{\sqrt{2}}\times \sin (x)=-\sqrt{2}\sin (x)...........(1.6)$
The right hand side of equation (1.6) is the same as the right hand side of the question. Hence, we have proved that
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin (x)$
Note: In this question, we did not expand the cosine terms separately but instead used equation (1.3) to evaluate the question. One can expand the cosine terms separately. However, then two terms involving the cosine terms will get cancelled out to give the same answer as we have obtained.
Complete step-by-step answer:
Here, we are given the cosines of two different angles and we have to find their sum. However, the angle in each cosine term is itself a sum or difference of two angles. Therefore, we should first understand the equation for finding the cosine of sum and difference of two angles.
We know that the expansion of cosine of sum and difference of two angles is given by
$\begin{align}
& \cos (a+b)=\cos (a)\cos (b)-\sin (a)\sin (b)........(1.1) \\
& \cos (a-b)=\cos (a)\cos (b)+\sin (a)\sin (b)............(1.2) \\
\end{align}$
If we subtract these two equations, we obtain
$\begin{align}
& \cos (a+b)-\cos (a-b)=\cos (a)\cos (b)-\sin (a)\sin (b)-\left( \cos (a)\cos (b)+\sin (a)\sin (b) \right) \\
& \Rightarrow \cos (a+b)-\cos (a-b)=-2\sin (a)\sin (b)....................(1.3) \\
\end{align}$
We find that this equation is similar to the expression in the question. Taking \[a=\dfrac{3\pi }{4}\]and \[b=x\] in equation (1.3), we obtain
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\sin \left( \dfrac{3\pi }{4} \right)\sin (x).......(1.4)$
Now, using the trigonometric equation
$\sin \left( \dfrac{\pi }{2}+x \right)=\sin (x)$
We obtain
\[\sin \left( \dfrac{3\pi }{4} \right)=\sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}.........(1.5)\]
Thus, using this value in equation (1.4), we obtain
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\times \dfrac{1}{\sqrt{2}}\times \sin (x)=-\sqrt{2}\sin (x)...........(1.6)$
The right hand side of equation (1.6) is the same as the right hand side of the question. Hence, we have proved that
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin (x)$
Note: In this question, we did not expand the cosine terms separately but instead used equation (1.3) to evaluate the question. One can expand the cosine terms separately. However, then two terms involving the cosine terms will get cancelled out to give the same answer as we have obtained.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: 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 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

