MHB What are the values of A and B for the given trigonometric expressions?

[Albert1]

find the following values:

$(1) A=cos^3\dfrac {\pi}{8}+cos^3\dfrac {3\pi}{8}+cos^3\dfrac {5\pi}{8}+cos^3\dfrac {7\pi}{8}=?$

$(2) B=cos^4\dfrac {\pi}{8}+cos^4\dfrac {3\pi}{8}+cos^4\dfrac {5\pi}{8}+cos^4\dfrac {7\pi}{8}=?$

 

[kaliprasad]

find the following values:

$(1) A=cos^3\dfrac {\pi}{8}+cos^3\dfrac {3\pi}{8}+cos^3\dfrac {5\pi}{8}+cos^3\dfrac {7\pi}{8}=?$

$(2) B=cos^4\dfrac {\pi}{8}+cos^4\dfrac {3\pi}{8}+cos^4\dfrac {5\pi}{8}+cos^4\dfrac {7\pi}{8}=?$

Two questions. let me find A first

Spoiler

as $ cos \dfrac{\pi}{8}=-\cos \dfrac{7\pi}{8}$

so $ cos^3 \dfrac{\pi}{8}=-\cos^3 \dfrac{7\pi}{8}$

hence $ cos^3 \dfrac{\pi}{8} +\cos^3 \dfrac{7\pi}{8}=0\cdots(1)$

similarly $ cos^3 \dfrac{3\pi}{8}+\cos^3 \dfrac{5\pi}{8}=0\cdots(2)$

adding above 2 we get
$ cos^3 \dfrac{\pi}{8}+\cos^3 \dfrac{3\pi}{8}+cos^3 \dfrac{5\pi}{8}+\cos^3 \dfrac{7\pi}{8} =0$

 

[kaliprasad]

now for B

Spoiler

we have
$ \cos(\dfrac{7\pi}{8}) = -\cos(\dfrac{\pi}{8})$
hence $ \cos ^4(\dfrac{7\pi}{8}) = \cos^4(\dfrac{\pi}{8})$
similarly
$ \cos(\dfrac{5\pi}{8}) = -\cos(\dfrac{3\pi}{8})$
hence $ \cos ^4(\dfrac{5\pi}{8}) = \cos^4(\dfrac{3\pi}{8})= \sin ^4 (\dfrac{\pi}{8})$

so we get
$ \cos^4 (\dfrac{\pi}{8}) + \cos^4(\dfrac{3\pi}{8})+ \cos^4(\dfrac{5\pi}{8})+ \cos^4(\dfrac{7\pi}{8}) $
= $2(\cos^4 (\dfrac{\pi}{8}) + \sin^4 (\dfrac{\pi}{8}))$
= $2(\cos^2 (\dfrac{\pi}{8}) + \sin^2 (\dfrac{\pi}{8}))^2- 2\cos^2 (\dfrac{\pi}{8})\sin^2 (\dfrac{\pi}{8}))$ using $a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2$
= $2 ( 1- \dfrac{1}{2} \sin ^2 \dfrac{\pi}{4})$
= $2 ( 1- \dfrac{1}{2} (\dfrac{1}{\sqrt{2}})^2)$
= $\dfrac{3}{2}$

 

[Albert1]

very good , both are correct