Question

For $A\in R$then prove that
1.$\sin A.\sin (60+A).\sin (60-A)=\dfrac{1}{4}\sin 3A$
2. $\cos A.\cos (60+A).\cos (60-A)=\dfrac{1}{4}\cos 3A$
3.$\sin 20{}^\circ .\sin 40{}^\circ .\sin 60{}^\circ .\sin 80{}^\circ =\dfrac{3}{16}$
4.$\cos \dfrac{\pi }{9}.\cos \dfrac{2\pi }{9}.\cos \dfrac{3\pi }{9}.\cos \dfrac{4\pi }{9}=\dfrac{1}{16}$

Answer 1

Hint: Basically for all four you have to be familiar with trigonometric identities. For first use$\sin (a+b)=\sin a\cos b+\cos a\sin b$and $\sin (a-b)=\sin a\cos b-\cos a\sin b$. For second use $\cos (A+B)=\cos A\cos B-\sin A\sin B$and$\cos (A-B)=\cos A\cos B+\sin A\sin B$. For third use $2\sin A\sin B=\cos (A-B)-\cos (A+B)$ . For fourth use $2\cos A\cos B=\cos (A+B)+\cos (A-B)$. Try it and you will get the answer.

Complete step-by-step 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 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 $\dfrac{\pi }{2}$).
For all, we have to prove LHS$=$RHS.
1.$\sin A.\sin (60+A).\sin (60-A)=\dfrac{1}{4}\sin 3A$

Now taking first one LHS$=\sin A.\sin (60+A).\sin (60-A)$
So we know $\sin (a+b)=\sin a\cos b+\cos a\sin b$and $\sin (a-b)=\sin a\cos b-\cos a\sin b$.
Applying above identities we get,
$=\sin A.(\sin 60\cos A+\cos 60\sin A).(\sin 60\cos A-\cos 60\sin A)$
Now we also know that ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$.
So applying above identity we get,
$=\sin A.({{\sin }^{2}}60{{\cos }^{2}}A-{{\cos }^{2}}60{{\sin }^{2}}A)$
$=\sin A.(\dfrac{3}{4}{{\cos }^{2}}A-\dfrac{1}{4}{{\sin }^{2}}A)$
So taking $\dfrac{1}{4}$ common we get,
$=\dfrac{1}{4}\sin A.(3{{\cos }^{2}}A-{{\sin }^{2}}A)$
$=\dfrac{1}{4}\sin A(3{{\cos }^{2}}A-{{\sin }^{2}}A)$
We know ${{\cos }^{2}}A=1-{{\sin }^{2}}A$.
So applying above identity we get,
\[\begin{align}
  & =\dfrac{1}{4}\sin A(3(1-{{\sin }^{2}}A)-{{\sin }^{2}}A) \\
 & =\dfrac{1}{4}\sin A(3-3{{\sin }^{2}}A-{{\sin }^{2}}A) \\
 & =\dfrac{1}{4}\sin A(3-4{{\sin }^{2}}A) \\
 & =\dfrac{1}{4}(3\sin A-4{{\sin }^{3}}A) \\
\end{align}\]
Here we are familiar with it \[(3\sin A-4{{\sin }^{3}}A)=\sin 3A\].
So we get,
\[=\dfrac{1}{4}\sin 3A=\]RHS
So we have proved LHS$=$RHS.
So $\sin A.\sin (60+A).\sin (60-A)=\dfrac{1}{4}\sin 3A$.
Hence proved.

2. $\cos A.\cos (60+A).\cos (60-A)=\dfrac{1}{4}\cos 3A$
For these 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 identity 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 $.
So applying above identity 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 identity 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.

3.$\sin 20{}^\circ .\sin 40{}^\circ .\sin 60{}^\circ .\sin 80{}^\circ =\dfrac{3}{16}$
So LHS$=\sin 20{}^\circ .\sin 40{}^\circ .\sin 60{}^\circ .\sin 80{}^\circ $
We have to use the product identity.
$2\sin A\sin B=\cos (A-B)-\cos (A+B)$
So applying above product identity we get,
$\begin{align}
  & =\dfrac{1}{2}(\sin 20{}^\circ .\sin 60{}^\circ )(2\sin 80{}^\circ .\sin 40{}^\circ ) \\
 & =\dfrac{1}{2}(\sin 20{}^\circ .\sin 60{}^\circ )(\cos (80-40)-\cos (80+40)) \\
 & =\dfrac{1}{2}(\dfrac{\sqrt{3}}{2}\sin 20{}^\circ )(\cos (40)-\cos (120)) \\
 & =\dfrac{\sqrt{3}}{4}\sin 20{}^\circ (\cos (40)-\cos (120)) \\
 & =\dfrac{\sqrt{3}}{4}\sin 20{}^\circ (\cos (40)+\dfrac{1}{2}) \\
\end{align}$…………$(\cos 60=\dfrac{1}{2}\And \cos 120=-\dfrac{1}{2})$
Now simplifying we get,
$=\dfrac{\sqrt{3}}{8}(2\sin 20{}^\circ \cos (40)+\sin 20)$
Now again using the product identity $2\sin A\cos B=\sin (A+B)+\sin (A-B)$.
$\begin{align}
  & =\dfrac{\sqrt{3}}{8}(\sin (20+40)+\sin (20-40)+\sin 20) \\
 & =\dfrac{\sqrt{3}}{8}(\sin (60)+\sin (-20)+\sin 20) \\
 & =\dfrac{\sqrt{3}}{8}(\sin (60)-\sin (20)+\sin 20) \\
 & =\dfrac{\sqrt{3}}{8}\sin (60) \\
\end{align}$
$=\dfrac{\sqrt{3}}{8}\times \dfrac{\sqrt{3}}{2}$
$=\dfrac{3}{16}=$RHS
So we have proved LHS$=$RHS.
$\sin 20{}^\circ .\sin 40{}^\circ .\sin 60{}^\circ .\sin 80{}^\circ =\dfrac{3}{16}$
Hence proved.

4.$\cos \dfrac{\pi }{9}.\cos \dfrac{2\pi }{9}.\cos \dfrac{3\pi }{9}.\cos \dfrac{4\pi }{9}=\dfrac{1}{16}$
Now LHS$=\cos \dfrac{\pi }{9}.\cos \dfrac{2\pi }{9}.\cos \dfrac{3\pi }{9}.\cos \dfrac{4\pi }{9}$
So now multiplying and dividing by $2$.
$=\dfrac{1}{2}\cos \dfrac{\pi }{3}.\cos \dfrac{4\pi }{9}(2\cos \dfrac{\pi }{9}.\cos \dfrac{2\pi }{9})$
So using $2\cos A\cos B=\cos (A+B)+\cos (A-B)$.
So applying above product identity we get,
$\begin{align}
  & =\dfrac{1}{2}\cos \dfrac{\pi }{3}.\cos \dfrac{4\pi }{9}(\cos (\dfrac{2\pi }{9}+\dfrac{\pi }{9})+\cos (\dfrac{2\pi }{9}-\dfrac{\pi }{9})) \\
 & =\dfrac{1}{2}\cos \dfrac{\pi }{3}.\cos \dfrac{4\pi }{9}(\cos (\dfrac{\pi }{3})+\cos (\dfrac{\pi }{9})) \\
 & =\dfrac{1}{2}.\dfrac{1}{2}.\cos \dfrac{4\pi }{9}(\dfrac{1}{2}+\cos (\dfrac{\pi }{9})) \\
\end{align}$……….$(\cos \dfrac{\pi }{3})=\dfrac{1}{2}$
$=\dfrac{1}{2}.\dfrac{1}{2}.(\cos \dfrac{4\pi }{9}\dfrac{1}{2}+\dfrac{1}{2}.2\cos \dfrac{4\pi }{9}\cos (\dfrac{\pi }{9}))$
Taking $\dfrac{1}{2}$ common we get,\
$=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.(\cos \dfrac{4\pi }{9}+2\cos \dfrac{4\pi }{9}\cos (\dfrac{\pi }{9}))$
Now again applying product identity we get,
$\begin{align}
  & =\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.(\cos \dfrac{4\pi }{9}+\cos (\dfrac{4\pi }{9}+\dfrac{\pi }{9})+\cos (\dfrac{4\pi }{9}-\dfrac{\pi }{9})) \\
 & =\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.(\cos \dfrac{4\pi }{9}+\cos (\dfrac{5\pi }{9})+\cos (\dfrac{\pi }{3})) \\
 & =\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.(\cos \dfrac{4\pi }{9}+\cos (\dfrac{5\pi }{9})+\dfrac{1}{2}) \\
\end{align}$

\[\begin{align}
  & =\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.(\cos \dfrac{4\pi }{9}+\cos (\pi -\dfrac{4\pi }{9})+\dfrac{1}{2}) \\
 & =\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.(\cos \dfrac{4\pi }{9}-\cos (\dfrac{4\pi }{9})+\dfrac{1}{2}) \\
 & =\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2} \\
\end{align}\]
$=\dfrac{1}{16}=$RHS
So we have proved LHS$=$RHS.
$\cos \dfrac{\pi }{9}.\cos \dfrac{2\pi }{9}.\cos \dfrac{3\pi }{9}.\cos \dfrac{4\pi }{9}=\dfrac{1}{16}$
Hence proved.

Note: You must know the concepts of trigonometric identities. You must know the product identity. Many students make mistakes in product identities. While simplifying be careful that no terms are to be missed. Also, while solving it, solve it easily.