Question

If cos Θ = negative two over five and tan Θ > 0, what is the value of sin Θ?

Answer 1

@ThereisnoUsername

Answer 2

\[a^2 + b^2 = c^2\]\[(-2)^2 + x^2 = 5^2\]\[x^2 = 25 - 4\]\[x = \sqrt{21}\]
\[\sin(Θ) = x/5\]\[\sin(Θ) = -\sqrt{21}/5\]

Answer 3

So, negative square root 21 over 5

Answer 4

wowww thank you !

Answer 5

how long are you going to be online?

Answer 6

you just use the pythagoren theorem
(a^2+b^2=c^2)

and probably all day I keep this page open while I work on my school work, I like doing the extra math and stuff it really helps XD

Answer 7

thats awesome i have more work to do as well so i will just tag you if anything .

Answer 8

@ThereisnoUsername ok. but please dont give directly right answers and try collaborate please with Asker to understanding the way how you solve same exercices right easy
hope you know the rules of OS.

thank you

Answer 9

right @ganeshie8 ?

Answer 10

@mathmate your opinion about above wrote ? ty.

Answer 11

@jhonyy9
Unfortunately I beg to differ from @ThereisnoUsername 's solution.

Answer 12

There are two ways to arrive at the same answer.
1. @ThereisnoUsername 's method (very brilliant, by the way).

so \(|sin(\theta)|=2/sqrt(2^2+5^2)\)
Considering that tan(theta) is negative, and theta>0, theta must fall in the 2nd quadrant.
So sin(theta) is positive (second quadrant).

2. use trigonometric identies
\(1+cot^2(\theta)=csc^2(\theta)\)
solving for sin(theta), we have
\(sin^2(\theta)=\frac{1}{1+cot^2(\theta)}=\frac{1}{1+\frac{1}{tan^2(\theta)}}\)
Evaluate sin^2(theta) and take square-root.
Determine sign of result as in part 1.

@ThereisnoUsername
In case you're wondering, \(tan(\theta)=opposite/adjacent\), so \((-2)^2+5^2=c^2\)