Question

I need help proving a trig identity. Question is in Attached photo

Answer 1

Here are the identity Given in the book

Answer 2

have you tried using sum identity for both terms on the left

Answer 3

sum/difference idenity

Answer 4

yes i end up getting 2(root2)cosx

Answer 5

for left side

Answer 6

ill show you what i did, maybe you can find my mistake

Answer 7

cool

Answer 8

i will gladly look at it

Answer 9

L.S=\[\sin(\frac{ \pi }{ 4 }+x) + \sin(\frac{ \pi }{ 4 }-x)\]
\[(\sin(\frac{ \pi }{ 4 })\cos x+\cos(\frac{ \pi }{ 4 }) \sin x)+(\sin(\frac{ \pi }{ 4 })\cos x- \cos(\frac{ \pi }{ 4 })\sin x)\]
\[=\sin(\frac{ \pi }{ 4 })\cos x+ \sin(\frac{ \pi }{ 4 })\cos x\]
\[=\sqrt{2}\cos x+\sqrt{2}\cos x\]
\[=2(\sqrt{2}\cos x)\]

Answer 10

oh wait i just realized sin (pi/4) does not equal root 2

Answer 11

\[\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\]

Answer 12

lol yeah

Answer 13

lol wow

Answer 14

disregard my last two line in my solution.
so then it would be
\[\frac{ \sqrt{2} }{ 2 }\cos x+\frac{ \sqrt{2} }{ 2 }\cos x\]
\[2\frac{ \sqrt{2} }{ 2 }\cos x\] <---the 2's cancel our and we are left with
\[\sqrt{2} \cos x\]
therefore L.S=R.S

Answer 15

totally correct

Answer 16

awsome

Answer 17

Thanks for the help :)

Answer 18

I didn't do much

Answer 19

You found your own error

Answer 20

Still when ever i post my question here, you guys keep me on the right tracks.

Answer 21

Trigonometry is not my cup of tea either, its the chapter i struggle with the most.

Answer 22

You will be a shark at it one day.

Answer 23

Also thanks to you guys I am getting a 95% in Math which is awsome