Question

Given sec(theta) = -4/3 and 90 degrees < (theta) <180 degrees ; find sin2(theta)

Answer 1

90 degrees < (theta) <180 degrees tells us that we are restricted in the 2nd quadrant... so all sine values are going to be positive...

Answer 2

If sec(theta) = -4/3
cos(theta) = -3/4

Use sin^2(x)+cos^2(x)=1 to solve for sin^2(x).
Plug in -3/4 for cos(x), square it, and subtract the result from 1.

Answer 3

\[\sin^2x\] or \[\sin2x\] for the last part of the question. They have different meanings.

Answer 4

\[\sin^2\theta............ \sin2\theta\]

Answer 5

its Sin2theta

Answer 6

alright... sin2theta has an identity
\[2\sin\theta \cos\theta\]

Answer 7

given that
sec(theta) = -4/3
sec(theta) =\[\frac{1}{-\frac{4}{3}} \rightarrow \frac{-3}{4}\]

Answer 8

it's negative due to the fact that we have the restriction of 90 degrees <theta<180 degrees
which means 2nd quadrant and only sine values are positive
now let's ignore the - for a bit and find out what our sine really is
so cosine is adjacent/hypotenuse
so that means cosine 3/4

Answer 9

so we need to use the pythagorean theorem a^2+b^2=c^2

Answer 10

in this case we solve for a.
a^2+3^2=4^2
a^2+9=16
a^2 = 16-9
a^2 = 7
a = \[\sqrt{7}\]
ugh one problem... I know we can't pick the negative result for a because negative sides are nonexistent and doesn't make sense... I don't think decimals can be used either... not sure.

Answer 11

sec(theta) = -4/3
cos(theta) = -3/4
in 2nd quadrant

hmmm... cos(theta) 3/4 = adjacent/hypotenuse
and then solve for a... that should be ...why the heck do I have square root of 7 ?
sine theta is \[\frac{\sqrt{7}}{4}\]
... I thought all sides are supposed to have whole numbers

Answer 12

@jim_thompson5910

Answer 13

It's perfectly possible to have a side length that isn't a whole number

example: