anonymous
  • anonymous
prove. sin [(pi over 4)+x] + sin [(pi over 4)-x] = sqrt{2} cosx
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

myininaya
  • myininaya
i think expanding both of those things might help
anonymous
  • anonymous
use ... sin(a=b)= sin(a)cos(b) +cos(a)sin(b) and sin(a-b)= sin(a)cos(b) -cos(a)sin(b)
anonymous
  • anonymous
all i know is that i will have to use addition indentities.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

myininaya
  • myininaya
\[[\sin(\frac{\pi}{4})\cos(x)+\sin(x)\cos(\frac{\pi}{4})]+[\sin(\frac{\pi}{4})\cos(x)-\sin(x)\cos(\frac{\pi}{4})]\]
myininaya
  • myininaya
\[\frac{\sqrt{2}}{2} \cos(x)+\sin(x) \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\cos(x)-\sin(x)\frac{\sqrt{2}}{2}\]
myininaya
  • myininaya
combine like terms and you will be done
anonymous
  • anonymous
its 2sin(a)cos(b)....now plug valus.....(2/\[\sqrt{2}\] ) sinx
anonymous
  • anonymous
\[\sqrt{2}\] sin x
anonymous
  • anonymous
you rock myininaya!!! thank you so much!!! and i appreciate your work and help Sam_unleashed! :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.