OpenStudy (anonymous):

1+cosx=2sin^2x Is there an identity I could substitute for?

2 years ago
OpenStudy (freckles):

do you know an identity for sin^2(x) that might be useful? check out the half identities under the half-angle identities http://www.purplemath.com/modules/idents.htm

2 years ago
OpenStudy (freckles):

oh wait are you solving an equation

2 years ago
OpenStudy (freckles):

or showing this is an identity or trying to show it isn't an identity

2 years ago
OpenStudy (freckles):

if this an equation you are solving just use sin^2(x)=1-cos^2(x)

2 years ago
OpenStudy (anonymous):

Solving the equation

2 years ago
OpenStudy (freckles):

then write everything on one side and realize you have a quadratic in terms of cosine

2 years ago
OpenStudy (freckles):

do you see that or not see this so far?

2 years ago
OpenStudy (freckles):

\[\text{ Replace } sin^2(x) \text{ with } 1-cos^2(x) \\ 1+cos(x)=2(1-cos^2(x)) \\ \text{ put everything on one side } \\ 1+cos(x)-2(1-cos^2(x))=0 \\ \text{ distribute a little } \]

2 years ago
OpenStudy (anonymous):

I don't see that

2 years ago
OpenStudy (freckles):

you don't see what?

2 years ago
OpenStudy (freckles):

the highest power is 2 and all the bases are cos(x)

2 years ago
OpenStudy (freckles):

so you have a quadratic in terms of cos

2 years ago
OpenStudy (freckles):

replace cos(x) with u might make it easier for you to see

2 years ago
OpenStudy (freckles):

\[1+u-2(1-u^2)=0\]

2 years ago
OpenStudy (anonymous):

Oh okay then. I see what you're saying.

2 years ago