Question

The value of ${{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}$ is
(A) $-1$
(B) $0$
(C) $1$
(D) $2i$

Answer 1

Hint: We need to find the value of ${{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}$ for answering this question. We will use the values ${{i}^{2}}=-1$ and the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and simplify the answer and come to a conclusion.

Complete step by step answer:
Now from the question we consider the expression ${{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}$ and we will simplify it and it reduces it we will have ${{\left( \left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right) \right)}^{8}}$ .
After performing the multiplication and simplifying it we will have ${{\left( {{\sin }^{2}}\dfrac{\pi }{8}-{{i}^{2}}{{\cos }^{2}}\dfrac{\pi }{8} \right)}^{8}}$ .
By using ${{i}^{2}}=-1$ and simplifying the expression as ${{\left( {{\sin }^{2}}\dfrac{\pi }{8}+{{\cos }^{2}}\dfrac{\pi }{8} \right)}^{8}}$.
As we know the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ we will use it here and simply write it as ${{\left( 1 \right)}^{8}}$ .
By calculating the further value we will have a conclusion that the simplified answer is $1$ .
Hence we come to a conclusion that the value of the expression ${{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}=1$ .
This is valid for any value of $\theta $.

So, the correct answer is “Option C”.

Note: While answering questions of this type we have to be careful with the simplifications and calculations and we should be sure about the trigonometric identities where to use what and remember them. There are mainly 3 trigonometric identities as follows: ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and ${{\sec }^{2}}x={{\tan }^{2}}x+1$ and ${{\csc }^{2}}x=1+{{\cot }^{2}}x$. While substituting the values we should take care that we substitute correct values if in case in place of ${{i}^{2}}=-1$ we had taken ${{i}^{2}}=1$ we will end up having a complete wrong conclusion as ${{\left( \sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8} \right)}^{8}}{{\left( \sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8} \right)}^{8}}={{\left( {{\sin }^{2}}\dfrac{\pi }{8}-{{\cos }^{2}}\dfrac{\pi }{8} \right)}^{8}}$ which would leave us in confusion.