How do you prove $(1+cosx)(1-cosx)=sin^2x$ ?

Question

31110 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-14T21:04:01

$sin^2x + cos^2x = 1$

the identity known is $sin^2x + cos^2x = 1$ .

this can be rearranged to give $1 - cos^2x = sin^2x$ .

using the 'difference of two squares' identity,

where $(a+b)(a-b) = a^2-b^2$ ,

$(1+cosx)(1-cosx) = 1^2 - cos^2x$

$1^2 = 1$

$(1+cosx)(1-cosx) = 1 - cos^2x$

since $1 - cos^2x = sin^2x$ , $(1+cosx)(1-cosx) = sin^2x$ .

Answer 2 · 2018-03-14T21:06:37

FOIL the LHS and use the trig identity that $sin^2x+cos^2x=1$

First, FOIL the left hand side (LHS):

$(1+cosx)(1-cosx)=1cancel(-cosx+cosx)+cos^2x$

$=1-cos^2x$

Now, re-write the equation:

$1-cos^2x=sin^2x$

There is a trig identity that states:

$sin^2x+cos^2x=1$

Substitute the 1 in our proof:

$sin^2xcancel(+cos^2x-cos^2x)=sin^2x$

$sin^2x=sin^2x$

How do you prove (1+cosx)(1-cosx)=sin^2x(1+cosx)(1-cosx)=sin^2x?