Question

Let $8(1+\cos \frac{\pi }{8})(1+\cos \frac{3\pi }{8})(1+\cos \frac{5\pi }{8})(1+\cos \frac{7\pi }{8})=\cos A.\cos B.\cos C.\cos D$ with $-\frac{\pi }{2}\le A,B,C,D\le \frac{\pi }{2}$. Then, which of the following is/are true?

• A. $A=B=C=D$
• B. $A+B=C+D$
• C. $A+B+C+D=0$
• D. $A-B=C-D$

Answer 1

The correct option is A $A=B=C=D$

Trigonometric identities

$\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}$

$\sin C-\sin D=2cos\frac{C+D}{2}\sin \frac{C-D}{2}$

$\cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}$

$\cos C-\cos D=2\sin \frac{C+D}{2}\sin \frac{D-C}{2}$

$\cos \frac{5\pi }{8}=\cos (\pi -\frac{3\pi }{8})=-\cos \frac{3\pi }{8}$

Also, $\cos \frac{7\pi }{8}=\cos (\pi -\frac{\pi }{8})=-\cos \frac{\pi }{8}$

$L.H.S=8(1+\cos \frac{\pi }{8})$$(1+\cos \frac{3\pi }{8})$$(1-\cos \frac{3\pi }{8})$$(1-\cos \frac{\pi }{8})$

$=8(1-{\cos }^2\frac{\pi }{8})$$(1-{\cos }^2\frac{3\pi }{8})$

$=8\left({\sin }^2\frac{\pi }{8}\right)$$\left({\sin }^2\frac{3\pi }{8}\right)$

$=2{\left(2\sin \frac{\pi }{8}\sin \frac{3\pi }{8}\right)}^2$

Using the identity above, we get

$L.H.S=2{(\cos \frac{\pi }{4}-\cos \frac{\pi }{2})}^2$

$=2{(\frac{1}{\sqrt{2}}-0)}^2$

$=1$

Now the R.H.S of the eqution given in the question can be $1$ only when $A=B=C=D=0$ for $-\frac{\pi }{2}\le A,B,C,D\le \frac{\pi }{2}$

Hence the correct answer is option $A$