Question

Prove that $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin x.$

Answer 1

We will use the trigonometric identity $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$ and the trigonometric identity $\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y.$ We know that $\sin \left( \pi -x \right)=\sin x.$ By using these trigonometric identities, we will expand the left-hand side of the given trigonometric identity.

Complete step by step solution:
Let us consider the given trigonometric identity $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin x.$
We are asked to prove the identity.
For that, we will first consider the left-hand side of the given trigonometric identity. Then we will expand each term of the identity using the trigonometric identities $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$ and $\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y.$
Let us consider the left-hand side of the given identity $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right).$
Now, let us expand the term $\cos \left( \dfrac{3\pi }{4}+x \right)$ using the trigonometric identity $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y.$
So, we will get $\cos \left( \dfrac{3\pi }{4}+x \right)=\cos \dfrac{3\pi }{4}\cos x-\sin \dfrac{3\pi }{4}\sin x.$
Similarly, we will expand the term $\cos \left( \dfrac{3\pi }{4}-x \right)$ using the trigonometric identity $\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y.$
We will get $\cos \left( \dfrac{3\pi }{4}-x \right)=\cos \dfrac{3\pi }{4}\cos x+\sin \dfrac{3\pi }{4}\sin x.$
Let us rewrite the left-hand side of the given identity using the expanded forms.
We will get,
$\Rightarrow \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=\cos \dfrac{3\pi }{4}\cos x-\sin \dfrac{3\pi }{4}\sin x-\left( \cos \dfrac{3\pi }{4}\cos x+\sin \dfrac{3\pi }{4}\sin x \right).$
From this, we will get the following when we open the brackets,
$\Rightarrow \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=\cos \dfrac{3\pi }{4}\cos x-\sin \dfrac{3\pi }{4}\sin x-\cos \dfrac{3\pi }{4}\cos x-\sin \dfrac{3\pi }{4}\sin x.$
Now, from this, we can cancel the similar terms with opposite signs.
Then as a result, we will get
$\Rightarrow \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sin \dfrac{3\pi }{4}\sin x-\sin \dfrac{3\pi }{4}\sin x.$
And we can write this as
$\Rightarrow \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\sin \dfrac{3\pi }{4}\sin x.$
Now, we know that $\dfrac{3\pi }{4}=\pi -\dfrac{\pi }{4}.$ Also, we know that the Sine function is positive in the second quadrant.
So, we will get $\sin \dfrac{3\pi }{4}=\sin \left( \pi -\dfrac{\pi }{4} \right).$
And, we will get $\sin \dfrac{3\pi }{4}=\sin \dfrac{\pi }{4}.$
Let us substitute this in the obtained equation.
We will get $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\sin \dfrac{\pi }{4}\sin x.$
We know that $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}.$
Therefore, we will get $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\dfrac{1}{\sqrt{2}}\sin x.$
We know that $\sqrt{2}\times \sqrt{2}=2.$ Therefore, $\dfrac{2}{\sqrt{2}}=\sqrt{2}.$
Thus, we will get $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin x.$
Hence the given identity is proved.

Note: Remember that all the trigonometric functions are positive in the first quadrant. The Sine function and thus the Cosecant function are positive in the second quadrant. The Tangent function and the Cotangent function are positive in the third quadrant. Similarly, the Cosine function and the Secant function are positive in the fourth quadrant.