Derive trigonometric functions

[mattmicko]

You should know the following trigonometric identities.
(A) sin(-x) = -sin x (C) cos(x + y) = cos x cos y - sin x sin y
(B) cos(-x) = cos x (D) sin(x + y) = sin x cos y + cos x sin y
Use these to derive the following important identities, which you should also know.
(a) sin^2(x) + cos^2(x) = 1 (use C and cos 0 = 1)

so i got to cos(x+y) = cos x cos y - sinx siny
then y=x so
cos(x+x)= cos^2 x - sin^2 x

that is where i'm stuck on that one.

 

[BigGlenntheHeavy]

\(\displaystyle What \ are \ you \ stuck \ on? \ Please \ be \ more \ explicit.\)

 

[mattmicko]

Sorry :/

Where i'm stuck is
cos(x+x)= cos^2 x - sin^2 x
because i don't know how to get the sin^2 x to be positive.
the closest i can get is
cos(0+0)=cos^2 x - sin^2 x
1= cos^2 x -sin^2 x

and i'm not even sure if that is correct.

 

[BigGlenntheHeavy]

\(\displaystyle Sorry, \ have \ no \ idea \ where \ you \ are \ coming \ from, \ perhaps \ someone \ else \ does.\)

 

[Deleted member 4993]

You should know the following trigonometric identities.
(A) sin(-x) = -sin x (C) cos(x + y) = cos x cos y - sin x sin y
(B) cos(-x) = cos x (D) sin(x + y) = sin x cos y + cos x sin y
Use these to derive the following important identities, which you should also know.
(a) sin^2(x) + cos^2(x) = 1 (use C and cos 0 = 1)

so i got to cos(x+y) = cos x cos y - sinx siny
then y=x so
cos(x+x)= cos^2 x - sin^2 x

that is where i'm stuck on that one.

Instead of y = x use

y = -x

 

[mattmicko]

Thank you very much. I've been working on this packet for the last three weeks, and I couldn't solve this one not matter what i did.

 

[Deleted member 4993]

Did you solve it now?

 

[mattmicko]

Yes I solved it thank you again.