Question

Write Out The Following Trigonometric Idenetites

Pythagorean - There are 3
Cos2x - There are 3
Sin2x

I never really learned these (was absent).

Answer 1

\[\sin ^{2}x+\cos ^{2}x=1\]
\[\cos2x=\cos ^{2}x-\sin^{2}x\]
\[\sin2x=2sinxcosx\]

Answer 2

Is that it?

Answer 3

others are
sec^2 x = 1 + tan^2 x
cosec^2 x = 1 + cot^2 x

(part 1)

Answer 4

Part 2?

Answer 5

cos 2x = cos^2 x - sin^2 x

the other 2 can be found by substituting
for cos^2 x and sin^2 x in the above

hint: use the first identity that myko posted

Answer 6

Elaborate a little further, please. I'm missing a lot of this information.

Answer 7

So far, I've got cos2x = 1-2sin^2 x

Answer 8

right thats one of them
the other one is obtained by substituting sin^2x = 1 - cos^2 x into the formula for cos 2x

Answer 9

So would I end up with an equation of cos2x = cos^2x - ( 1 - cos^2 x)
?
If so, how would I simplify it? Or is that the answer?

Answer 10

cos2x = cos^2x - ( 1 - cos^2 x)
= cos^2x - 1 + cos^2 x ( sign changes when parentheses are removed)
= 2cos^2 x - 1

Answer 11

Glad to see I was on the right track. Thanks for the help.

Answer 12

yw