Question

# Solve $\cos A.\cos (60+A).\cos (60-A)=\dfrac{1}{4}\cos 3A$.

Hint: Take LHS and simplify it. While simplifying, use the identities $\cos (A+B)=\cos A\cos B-\sin A\sin B$,$\cos (A-B)=\cos A\cos B+\sin A\sin B$,${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$ and $\cos 3A=4{{\cos }^{3}}A-3\cos A$. You will get the answer.

The trigonometric functions (also called circular functions, angle functions, or trigonometric functions) are real functions that relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. They are among the simplest periodic functions, and as such are also widely used for studying periodic phenomena, through Fourier analysis.
The most widely used trigonometric functions are the sine, the cosine, and the tangent. Their reciprocals are respectively the cosecant, the secant, and the cotangent, which are less used in modern Mathematics.

The cosine function, along with sine and tangent, is one of the three most common trigonometric functions. In any right triangle, the cosine of an angle is the length of the adjacent side (A) divided by the length of the hypotenuse (H). In a formula, it is written simply as '$\cos$'. $\cos$ function (or cosine function) in a triangle is the ratio of the adjacent side to that of the hypotenuse. The cosine function is one of the three main primary trigonometric functions and it is itself the complement of sine (co + sine).
The cosine graph or the cos graph is an up-down graph just like the sine graph. The only difference between the sine graph and the cos graph is that the sine graph starts from $0$ while the cos graph starts from $90{}^\circ$ (or $\dfrac{\pi }{2}$).
So we have to prove LHS$=$RHS.
Now taking LHS$=\cos A.\cos (60+A).\cos (60-A)$, we know the identities $\cos (A+B)=\cos A\cos B-\sin A\sin B$ and $\cos (A-B)=\cos A\cos B+\sin A\sin B$.
Applying the above identities we get,
$=\cos A.(\cos 60\cos A-\sin 60\sin A).(\cos 60\cos A+\sin 60\sin A)$
Now we know ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$.
So applying above we get,
\begin{align} & =\cos A.({{\cos }^{2}}60{{\cos }^{2}}A-{{\sin }^{2}}60{{\sin }^{2}}A) \\ & =\cos A.(\dfrac{1}{4}{{\cos }^{2}}A-\dfrac{3}{4}{{\sin }^{2}}A) \\ \end{align}………….. ($\cos 60=\dfrac{1}{2}$ and $\sin 60=\dfrac{\sqrt{3}}{2}$)
Taking $\dfrac{1}{4}$ common we get,
$=\dfrac{1}{4}\cos A.({{\cos }^{2}}A-3{{\sin }^{2}}A)$
We also know ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta$. Applying this property, we get,
\begin{align} & =\dfrac{1}{4}\cos A.({{\cos }^{2}}A-3(1-{{\cos }^{2}}A)) \\ & =\dfrac{1}{4}\cos A.({{\cos }^{2}}A-3+3{{\cos }^{2}}A) \\ & =\dfrac{1}{4}\cos A.(4{{\cos }^{2}}A-3) \\ & =\dfrac{1}{4}(4{{\cos }^{3}}A-3\cos A) \\ \end{align}
Here we know the identity $\cos 3A=4{{\cos }^{3}}A-3\cos A$.
So applying above we get,
$=\dfrac{1}{4}\cos 3A=$RHS
So we have proved LHS$=$RHS.
$\cos A.\cos (60+A).\cos (60-A)=\dfrac{1}{4}\cos 3A$
Hence proved.

Note: Carefully read the question. Here to prove LHS$=$RHS you should be familiar with the identities. Most of the mistakes are done while simplifying so kindly avoid the mistakes while simplifying. Try to see what is the major concept behind these.