Question

How would you verify the following identity \[\dfrac{\cos \left( 3x \right)}{\cos x}=1-4{{\sin }^{2}}x?\]

Answer 1

Hint: We are asked to verify that \[\dfrac{\cos 3x}{\cos x}\] is the same as \[1-4{{\sin }^{2}}x.\] To do so we will use the sum formula of cos x then we use cos (A + B) = cos A cos B – sin A sin B. We will also use sin (2x) = 2 sin x cos x. Then we will also use the identity \[{{\cos }^{2}}x+{{\sin }^{2}}x=1.\] Using these we will start with the left side and will move forward to the right side. Once we get the right side, our problem is solved.

Complete step by step answer:
We are asked to verify that \[\dfrac{\cos 3x}{\cos x}=1-4{{\sin }^{2}}x.\] Now we start our solution by considering the left-hand side. So, we have \[\dfrac{\cos 3x}{\cos }.\] As we can see that 3x = 2x + x, so we get,
\[\dfrac{\cos 3x}{\cos x}=\dfrac{\cos \left( 2x-1x \right)}{\cos x}.....\left( i \right)\]
Now, as we know that cos (A + B) is given as \[\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B,\] so we get,
\[\cos \left( 2x+x \right)=\cos \left( 2x \right)\cos \left( x \right)-\sin \left( 2x \right)\sin x\]
Using in (i), we get,
\[\Rightarrow \dfrac{\cos 3x}{\cos x}=\dfrac{\cos 2x\cos x-\sin 2x\sin x}{\cos x}\]
Now, as we know that \[\sin 2x=2\sin x\cos x,\] so using this, we get,
\[\Rightarrow \dfrac{\cos 3x}{\cos x}=\dfrac{\cos 2x\cos x-2\sin x\cos x\sin x}{\cos x}\]
As cos x is common in the numerator in both the terms, so we take it out. So, we get,
\[\Rightarrow \dfrac{\cos 3x}{\cos x}=\dfrac{\cos x\left[ \cos 2x-2{{\sin }^{2}}x \right]}{\cos x}\]
Now, cancelling the like terms, we get,
\[\Rightarrow \dfrac{\cos 3x}{\cos x}=\cos 2x-2{{\sin }^{2}}x\]
Now, as we know that cos 2 x is given as \[2{{\cos }^{2}}x-1\] and using \[{{\sin }^{2}}x+{{\cos }^{2}}x=1,\] we have \[{{\sin }^{2}}x=1-{{\cos }^{2}}x.\] So using these above, we get,
\[\Rightarrow \dfrac{\cos 3x}{\cos x}=\left( 2{{\cos }^{2}}x-1 \right)-2\left( 1-{{\cos }^{2}}x \right)\]
\[\Rightarrow \dfrac{\cos 3x}{\cos x}=2{{\cos }^{2}}x-1-2+2{{\cos }^{2}}x\]
On simplifying, we get,
\[\Rightarrow \dfrac{\cos 3x}{\cos x}=4{{\cos }^{2}}x-3\]
Now changing \[{{\cos }^{2}}x\] into \[{{\sin }^{2}}x\] using \[{{\cos }^{2}}x=1-{{\sin }^{2}}x,\] we will get,
\[\Rightarrow \dfrac{\cos 3x}{\cos x}=4\left( 1-{{\sin }^{2}}x \right)-3\]
\[\Rightarrow \dfrac{\cos 3x}{\cos x}=4-4{{\sin }^{2}}x-3\]
On simplifying, we get,
\[\Rightarrow \dfrac{\cos 3x}{\cos x}=1-4{{\sin }^{2}}x\]
So, we get LHS = RHS. Hence our solution is verified.

Note:
Remember that we cannot verify an equation using just one term that satisfies. If we have one term that does not satisfy the equation, then we can see that the equation is unequal but if one value is there which satisfies but we are not sure about then we cannot say they are the same. For example,
\[\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\]
\[\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}\]
But these two functions are different but they are the same at 45 degrees.