Question

simplify this expression

Answer 1

Also can someone Factor the algebraic expression below in terms of a single trigonometric function.

cos x - sin 2x - 1

Answer 2

for the first 1 look here
http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity

Answer 3

now in the second one open sin2x and see what you can take common

Answer 4

what?

Answer 5

for the 1st one, use the pythagorean identities -> http://web.mit.edu/wwmath/trig/eq1.gif

for the 2nd one, use the double-angle identities -> http://www.sosmath.com/trig/Trig5/trig5/img7.gif

Answer 6

for the 1st one, the numerator is simple, check the identities
the denominator... just use the cotangent pythagorean identity

for the 2nd one, use the sine double-angle identity

Answer 7

someone show me how to do the first one please

Answer 8

\(\bf \cfrac{cos^2(x)+sin^2(x)}{cot^2(x)-csc^2(x)}\implies \cfrac{\bbox[border: 1px solid black]{ \textit{check the identities, what do you think?} } }{cot^2(x)-csc^2(x)}\)

Answer 9

im so confused

Answer 10

well... do you see the pythagorean identities?

Answer 11

csc x

-1

1

sec x

Answer 12

those are the answers and im confused

Answer 13

well... yes... so... which one can we use in the numerator from -> http://web.mit.edu/wwmath/trig/eq1.gif ?

Answer 14

so the top is one but how do you figure out the bottom is it cscx?

Answer 15

\(\bf \cfrac{cos^2(x)+sin^2(x)}{cot^2(x)-csc^2(x)}\implies \cfrac{1}{cot^2(x)-csc^2(x)}\\ \quad \\
\textit{recall }1+cot^2(x)=csc^2(x)\implies cot^2(x)=csc^2(x)-1\qquad thus\\ \quad \\
\cfrac{1}{cot^2(x)-csc^2(x)}\implies
\cfrac{1}{[\cancel{csc^2(x)}-1]\cancel{-csc^2(x)}}\)

Answer 16

so the answer is -1

Answer 17

yes