Sin 8/ 1+ cos 8 = 1-
cos 8 /1+ cos 8
Answers
Answer:
Answer:
\left( 1 + \cos \frac { \pi } { 8 } \right) \left( 1 + \cos \frac { 3 \pi } { 8 } \right) \left( 1 + \cos \frac { 5 \pi } { 8 } \right) \left( 1 + \cos \frac { 7 \pi } { 8 } \right) = \frac { 1 } { 8 }(1+cos
8
π
)(1+cos
8
3π
)(1+cos
8
5π
)(1+cos
8
7π
)=
8
1
Solution:
Given that
\begin{lgathered}\begin{array} { l } { \left( 1 + \cos \frac { \pi } { 8 } \right) \left( 1 + \cos \frac { 3 \pi } { 8 } \right) \left( 1 + \cos \frac { 5 \pi } { 8 } \right) \left( 1 + \cos \frac { 7 \pi } { 8 } \right) } \\\\ { = \left( 1 + \cos \frac { \pi } { 8 } \right) \left( 1 + \sin \left( \frac { \pi } { 2 } - \frac { 3 \pi } { 8 } \right) \right) \left( 1 + \sin \left( \frac { \pi } { 2 } - \frac { 5 \pi } { 8 } \right) \right) \left( 1 + \cos \left( \pi - \frac { 7 \pi } { 8 } \right) \right) } \end{array}\end{lgathered}
(1+cos
8
π
)(1+cos
8
3π
)(1+cos
8
5π
)(1+cos
8
7π
)
=(1+cos
8
π
)(1+sin(
2
π
−
8
3π
))(1+sin(
2
π
−
8
5π
))(1+cos(π−
8
7π
))
\because \cos \left( \frac { \pi } { 2 } - X \right) = \sin X∵cos(
2
π
−X)=sinX
\begin{lgathered}\begin{array} { l } { = \left( 1 + \cos \frac { \pi } { 8 } \right) \left( 1 + \sin \left( \frac { \pi } { 8 } \right) \right) \left( 1 - \sin \left( \frac { \pi } { 8 } \right) \right) \left( 1 - \cos \left( \frac { \pi } { 8 } \right) \right) } \\\\ { = \left( 1 - \cos ^ { 2 } \frac { \pi } { 8 } \right) \left( 1 - \sin ^ { 2 } \left( \frac { \pi } { 8 } \right) \right) } \end{array}\end{lgathered}
=(1+cos
8
π
)(1+sin(
8
π
))(1−sin(
8
π
))(1−cos(
8
π
))
=(1−cos
2
8
π
)(1−sin
2
(
8
π
))
\begin{lgathered}\begin{array} { l } { = \sin ^ { 2 } \left( \frac { \pi } { 8 } \right) \cos ^ { 2 } \left( \frac { \pi } { 8 } \right) } \\\\ { = \frac { 1 } { 4 } \times \left( 2 \sin \left( \frac { \pi } { 8 } \right) \cos \left( \frac { \pi } { 8 } \right) \right) ^ { 2 } } \end{array}\end{lgathered}
=sin
2
(
8
π
)cos
2
(
8
π
)
=
4
1
×(2sin(
8
π
)cos(
8
π
))
2
Using double angle formula,
\begin{lgathered}\begin{array} { l } { = \frac { 1 } { 4 } \times \sin ^ { 2 } \left( \frac { 2 \pi } { 8 } \right) = \frac { 1 } { 4 } \times \sin ^ { 2 } \left( \frac { \pi } { 4 } \right) } \\\\ { = \frac { 1 } { 4 } \times \left( \frac { 1 } { \sqrt { 2 } } \right) ^ { 2 } = \frac { 1 } { 8 } } \end{array}\end{lgathered}
=
4
1
×sin
2
(
8
2π
)=
4
1
×sin
2
(
4
π
)
=
4
1
×(
2
1
)
2
=
8
1
Hence,
\left( 1 + \cos \frac { \pi } { 8 } \right) \left( 1 + \cos \frac { 3 \pi } { 8 } \right) \left( 1 + \cos \frac { 5 \pi } { 8 } \right) \left( 1 + \cos \frac { 7 \pi } { 8 } \right) = \frac { 1 } { 8 }(1+cos
8
π
)(1+cos
8
3π
)(1+cos
8
5π