Question

cos^2(pi/8+x/2) - sin^2(pi/8-x/2) is equal to

Answer 1

use this : \(\large \cos(\theta) = \sin(\frac{\pi}{2} - \theta)\)

Answer 2

sin(pi/2-pi/8-x/2)

Answer 3

yes :)

\[\large \cos\left(\frac{\pi}{8} + \frac{x}{2}\right) =\sin\left(\frac{\pi}{2} - \left(\frac{\pi}{8} + \frac{x}{2}\right)\right) \]

Answer 4

simplify pi/2-pi/8

Answer 5

3pi/8

Answer 6

Ahh yeah it doesn't help :/

Answer 7

try double angle identities :
\[\large \cos^2 \theta = \dfrac{1+\cos2\theta }{2}\]
\[\large \sin^2 \theta = \dfrac{1-\cos2\theta }{2}\]

Answer 8

\[ \large \cos^2(\pi/8+x/2) - \sin^2(\pi/8-x/2)\]

\[ \large \dfrac{1+\cos(\pi/4+x)}{2} - \dfrac{1-\cos(\pi/4-x)}{2} \]

\[ \large \dfrac{1}{2}(~ \cos(\pi/4+x) + \cos(\pi/4-x) ~) \]

\[ \large \dfrac{1}{2}(~ 2\cos(\pi/4) \cos(x)~) \]

\[ \large\cos(\pi/4) \cos(x) \]

Answer 9

1/sqrt(2)cosx

Answer 10

yes @No.name

Answer 11

Is theta the same or different in both the cases would you check it please

Answer 12

different

Answer 13

oh o k