Question

I have never seen this before how do i go about doing it?

solve for x in terms of pi

sin(x + pi/4)+sin(x - pi/4)=-1

Answer 1

do i need to use the sin identities? sin(A)cos(b)+sin(B)cos(A) and sin(A)cos(b)-sin(B)cos(A)?

Answer 2

Yes...

Answer 3

thanks for that ill post the answer when i get it

Answer 4

:)

Answer 5

see sin(x + pi/4) +sin(x - pi/4)=(1/(v2))(sinx+cosx) +(1/(v2))(sinx-cosx)=V2 sinx

hence V2sinx=-1 or sinx=-1/sqrt(2) =
or sinx =-sin(pi/4) or sinx=sn(pi+pi/4) =sin(5pi/4)
hence x=5pi/4

Answer 6

Why can't you let him do it? @matricked

Answer 7

lol dude when helping people on this site @matricked you should rather let them do it themselves rather than providing the answer. There is no benefit in doing that.

Answer 8

you still have a chance to do it yourself @Razzputin ..matricked didn't use the trig formulas

Answer 9

yup i know im just asking him to refrain from doing what he does, when someone does that my eyes just drift over.

Answer 10

spoils the fun doesn't it

Answer 11

so i get to \[2\sin \frac{ \sqrt{2} }{ 2 }+2\cos \frac{ \sqrt{2} }{ 2 }=-1\] am i right so far?

Answer 12

\[sinx+cosx=\frac{ -1 }{ \sqrt{2} }\]

Answer 13

i think i did something wrong?

Answer 14

Yes..recheck.

Answer 15

\[sinx * \cos \frac{ \pi }{ 4 }+\sin\frac{ \pi }{ 4 } * cosx +sinx * \cos -\frac{ \pi }{ 4 }-\sin-\frac{ \pi }{ 4 } * cosx=-1\]

\[sinx *\frac{ \sqrt2 }{ 2 }+\frac{ \sqrt2 }{ 2 } * cosx +sinx * \frac{ \sqrt2 }{ 2 }--\frac{ \sqrt2 }{ 2 } * cosx=-1\]


\[\frac{ \sqrt2 }{ 2 }sinx +\frac{ \sqrt2 }{ 2 }cosx +\frac{ \sqrt2 }{ 2 }sinx+\frac{ \sqrt2 }{ 2 }cosx=-1\]


\[2*\frac{ \sqrt2 }{ 2 }sinx +2*\frac{ \sqrt2 }{ 2 }cosx=-1\]


\[ \sqrt2 sinx +\sqrt2 cosx=-1\]


\[ sinx +cosx=\frac{ -1 }{ \sqrt2 }\]


where did i go wrong?

Answer 16

\[\sin(x+\pi/4) = sinx.\cos \pi/4 + cosx.\sin \pi/4\]
\[\sin(x-\pi/4) = sinx.\cos \pi/4 - cosx.\sin \pi/4\]

Answer 17

You made a mistake in your first step. did you notice it now?

Answer 18

ahhh i see so the sign does not cary into the equation rather determines which equation?

Answer 19

Yes..

Answer 20

If you carry the sign into the equation, you have to use only the first equation...

Answer 21

so i need to have \[-1 = 2sinx \frac{\sqrt2}{2}\]

Answer 22

Yes..

Answer 23

so x = -45??

Answer 24

\[-\frac{1}{4}\pi\]

Answer 25

The general solution is \[(4n+3)\pi/2 \pm \pi/4\]

Answer 26

the exact answer depends on the interval given

Answer 27

its not right. i get 0 instead of -1 if i sub in\[ -\frac{1}{4}\pi\] how do you get to the answer you got?

Answer 28

No If you substitute -45 in your question, you get -1...

Answer 29

but -45 is the same as /[-frac{1}{4}pi/]

Answer 30

the answer is \[\sin x = -1/\sqrt{2}\]

Answer 31

Hey wait...what are you saying? Let us be clear..What did you get and what is wrong?

Answer 32

Yes -45 is same as -pi/4

Answer 33

ok i got \[sinx = \frac{ -1 }{ \sqrt2 }\] which equals x = -45 in degrees and \[x= \frac{-1}{4}\pi\] in radians. When i substitute those two answers in i dont get -1

Answer 34

sin(-45+45) + sin(-45-45) = sin0 + sin(-90) = 0 + (-1) = -1

Answer 35

ok now type that in but instead use the values x = -pi/4 and pi/4

Answer 36

i get - 0.02741213359

Answer 37

thats if i use my calculator set to degrees once i set it to radians it gives me -1...

Answer 38

thank you for your help @AbhimanyuPudi

Answer 39

welcome:)

Answer 40

cool, i learnt two things in one here... never mix degrees and radians on a calculator it isnt clever enough to determine that and also the trick you showed me with the two equations