Question

1-2tan^2xcos^2=2cosx-1 prove te identity

Answer 1

hmmm... is there a square in the right-hand side? \(\bf 1-tan^2(x)cos^2(x)=2cos^2(x)-1\quad ?\)

Answer 2

hmm missing the 2... \(\bf 1-2tan^2(x)cos^2(x)=2cos^2(x)-1\quad ?\)

Answer 3

yes there is
1-2tan^2xcos^2x=2cos^2x-1

Answer 4

hmm do you recall what tangent = ?

Answer 5

identity wise that is

Answer 6

no

Answer 7

http://www.mathwords.com/t/trig_identities.htm <--- can you see what tangent equals to?

Answer 8

ok thank you

Answer 9

\(\bf tan(x)=\cfrac{sin(x)}{cos(x)}\) thus \(\bf tan^2(x)=\cfrac{sin^2(x)}{cos^2(x)}\)

Answer 10

right i get that part , but how do i incorporate the 2 in front of
tan

Answer 11

\(\bf 1-2tan^2(x)cos^2(x)\implies
1-2\left(\cfrac{sin^2(x)}{\cancel{cos^2(x)}}\right)\cancel{cos^2(x)}\\ \quad \\\implies 1-2sin^2(x)\)

Answer 12

that's the left-side

now let's take a peek at the right-side \(\bf \textit{recall that}\quad sin^2(x)+cos^2(x)=1\implies cos^2(x)=1-sin^2(x)\quad thus\\ \quad \\
2cos^2(x)-1\implies 2[1-sin^2(x)]-1\)

Answer 13

so.... what would you get for the right-side ?

Answer 14

2cos^2(x)-1 right?

Answer 15

\(\bf 2cos^2(x)-1\implies 2[1-sin^2(x)]-1\) multiply the 2 in front of the bracket, and cancel out like-terms

Answer 16

ok i got it now thanks a lot you were extremely helpful

Answer 17

yw

Answer 18

i have another question that is giving me difficulty