Cos(45 - v) = sin (v + 45) for all angles v?

[Maria]

How can I prove that
cos(45 - v) = sin (v + 45) for all angles v?

 

[TenaliRaman]

expand LHS using cos(A-B) formula
expand RHS using sin(A+B) formula
show that they are equivalent

-- AI

 

[marlon]

or use the fact that cos(x) = sin(90°-x).

ofcourse if you want to prove the above relation you will have to follow to advice of TenaliRaman.

regards
marlon

 

[Maria]

can one of you show me? I don`t really knowwhere to begin? :shy:

 

[Zurtex]

You have, cos(45° - v) = sin (v + 45°)

Now as said before you should be aware of the relationship, cos(x) = sin(90°-x). All you have to do with this is let x = 45° - v.

However if you work is in context of the addition of angles then:

\sin (A \pm B) = \sin A \cos B \pm \sin B \cos A

\cos (A \pm B) = \cos A \cos B \mp \sin A \sin B

Let A = 45° and B = v.

 

[Maria]

so then I get:
sin(45+v) = sin 45 cos v + sin v cos 45
cos(45-v) = cos 45 cos v + sin 45 sin v

does this prove that cos(45-v) = sin(v+45)?

 

[Zurtex]

so then I get:
sin(45+v) = sin 45 cos v + sin v cos 45
cos(45-v) = cos 45 cos v + sin 45 sin v

does this prove that cos(45-v) = sin(v+45)?

Almost, what does cos 45° and sin 45° equal?

 

[Maria]

0,7071?

So I don`t have to write more that this?

I don`t really think I`ve got it yet..

 

[Zurtex]

0,7071?

So I don`t have to write more that this?

I don`t really think I`ve got it yet..

Correct me if I am wrong but both cos 45° and sin 45° are \frac{\sqrt{2}}{2}

Therefore:

\sin (45+v) = \frac{\sqrt{2}}{2} \cos v + \frac{\sqrt{2}}{2} \sin v
\cos (45-v) = \frac{\sqrt{2}}{2} \cos v + \frac{\sqrt{2}}{2} \sin v

Spot something simmilar? When proving things never ever ever ever ever ever ever ever ever ever ever ever round things off!

 

[Maria]

I didn`t know that.. thanks a lot..

 

[Galileo]

It's easier with sin(x) = cos(x-90).
cos(45-v)=cos(v-45) since the cosine is even.
cos(v-45)=sin(v+90-45)=sin(v+45)