Double-Angle Theorem: sec(@) = (-5root21)/21, csc(@)>0, find

[Barken]

The equation is

sec theta = (-5root21)/21 , csc theta > 0 Find sin(2theta).

I'm completely lost, but my idea is that you would have to use a cos(2theta) double angle formula, I'm not quite sure which one to use. Could someone please give me a hand explaining this to me. Thanks

 

[soroban]

Re: Double-Angle Theorem

Hello, Barken!

\(\displaystyle \sec\theta \:=\:\text{-}\frac{5\sqrt{21}}{21},\quad \csc\theta \,>\, 0\)

\(\displaystyle \text{Find }\,\sin(2\theta)\)

We are told that: .\(\displaystyle \sec\theta \:<\:0\)
. . Secant is negative in Quadrants 2 and 3.

We are told that: .\(\displaystyle \csc\theta \:>\:0\)
. . Cosecant is positive in Quadrants 1 and 2.

Hence, \(\displaystyle \theta\) is in Quadrant 2.


\(\displaystyle \text{We have: }\:\sec\theta \:=\:\text{-}\frac{5}{\sqrt{21}} \:=\:\frac{hyp}{adj}\)

\(\displaystyle \theta\text{ is an angle with: }\:adj = \text{-}\sqrt{21},\;hyp = 5\)

\(\displaystyle \text{Pythagorus says: }\pp \:=\:\pm2\)

\(\displaystyle \text{Since }\theta\text{ is in Quadrant 2, }\pp \:=\:+2\)

\(\displaystyle \text{Hence: }\;\sin\theta \:=\:\frac{2}{5},\quad \cos\theta \:=\:-\frac{\sqrt{21}}{5}\)


\(\displaystyle \text{Therefore: }\;\sin(2\theta) \;=\;2\sin\theta\cos\theta \;=\;2\left(\frac{2}{5}\right)\left(-\frac{\sqrt{21}}{5}\right) \;=\;-\frac{4\sqrt{21}}{25}\)

 

[Deleted member 4993]

Re: Double-Angle Theorem

The equation is

sec theta = (-5root21)/21 , csc theta > 0 Find sin(2theta).

I'm completely lost, but my idea is that you would have to use a cos(2theta) double angle formula, I'm not quite sure which one to use. Could someone please give me a hand explaining this to me. Thanks

Another way

sec(?) = -5?(21)/21

sec[sup:m2qtfclq]2[/sup:m2qtfclq](?) = 25/21

1 + tan[sup:m2qtfclq]2[/sup:m2qtfclq](?) = 25/21

tan[sup:m2qtfclq]2[/sup:m2qtfclq](?) = 4/21

tan(?) = - 2/?(21)

sin(?)•cos(?) = tan(?)/sec[sup:m2qtfclq]2[/sup:m2qtfclq](?) = [- 2/?(21)]/(25/21) = -2?(21)/25

sin(2?) = 2sin(?)•cos(?) = -4?(21)/25

You can do this yet another way - using cos(2?) - I'll leave that to you to find it.