Question

sin(x+pi/4)=sin(x−pi/4)−1
my answer is 3pi/4 and 5pi/4. i don't think it's correct, can somebody help me please?

Answer 1

I think I might try using the formulas for sin(x+y) and sin(x-y). Have you tried that?

Answer 2

yeah, that's how i came up with those answers

Answer 3

the sum and diference identities, right?

Answer 4

\[\sin x \cos 45+\cos x \sin 45=\sin x \cos 45-\cos x \sin 45-1=\frac{\sqrt{2}}{2}(\sin x+\cos x)=\frac{\sqrt{2}}{2}(\sin x-\cos x)-1\]

Answer 5

yes

Answer 6

umm...it's cut off

Answer 7

So we have
\[2\cos x \sin 45=1\]

Answer 8

\[2(\frac{\sqrt{2}}{2})\cos x=1\]

Answer 9

\[\cos x=\frac{\sqrt{2}}{2}\]

Answer 10

wait, it's not = -1?

Answer 11

i thought it was = -1, not 1

Answer 12

\[x=\frac{3pi}{4}, \frac{5pi}{4}\]

Answer 13

Yep. You are correct. negative 1

Answer 14

Oh no. I had to go and make that mistake when satellite is watching.

Answer 15

so there are two answers?

Answer 16

lol

Answer 17

i think there is also a way to write
\[\sin(x+\frac{\pi}{4})-\sin(x-\frac{\pi}{4})\] as a single function of cosine

Answer 18

yes. two answers between 0 and 2pi. Infinite number of solutions if you want them all.

Answer 19

nvm that is what you did. i guess i mean another way but it amounts to the same thing

Answer 20

yup, it's only in the interval of 0<x<2pi

Answer 21

I think you are correct but who wanted to look that up.

Answer 22

great, Thanks!

Answer 23

yw

Answer 24

yeah you can turn
\(-\sin(x-\frac{\pi}{4})\) in to \(\cos(x+\frac{\pi}{4})\) then
\[\sin(x+\frac{\pi}{4})+\cos(x+\frac{\pi}{4})=\sqrt{2}\cos(x)\] or something like that
i always forget how to do these
myininaya remembers