One More Trig Identity Problem

[DethRose]

There is one last problem i have on my trig assignment and i have no clue how to do it. The questions is:

Find sinx, cosx, tanx, sin2x, cos2x, and tan2x from the given information:

secx=5, sin is negative.

If anyone could show me how to do this problem it would be soooo appreciated.

Thanks

 

[courtrigrad]

$$\sec x = 5 \Rightarrow \frac{1}{\sec x} = \frac{1}{5} = \cos x$$

$$\cos x$$ is positive and $$\sin x$$ is negative in the fourth quadrant.

$$\sin 2x = 2\sin x \cos x$$

$$\cos 2x = 1-2\sin^{2} x$$

 

[DethRose]

so how do i find sinx cause if you go from cos wouldn't you have to express sin in terms of tan since cos=sin/tan?

 

[courtrigrad]

draw a right triangle. If you know the adjacent side is 1, and the hypotenuse is 5, then the opposite side is $$2\sqrt{6}$$.

 

[DethRose]

I don't understand how that will help?

 

[DethRose]

not that I am saying your wrong lol I am just saying i don't understand how to do that

 

[DethRose]

ah i get you

 

[DethRose]

wouldnt the other side be square root of 24 if you use pythagorins theorem?

 

[courtrigrad]

yes it would. sorry, a typo.

 

[DethRose]

Ok just to make sure I am doing it correctly here are the answers i got, would be great if you could tell me if i totally messed it up lol.

I got:

sinx= -2 squareroot 6/5
tanx= -2 squareroot 6
sin2x= -4 squareroot 6/25
cos2x= 23/25
tan2x= -4 squareroot 6/5

thanks for all the help!

 

[courtrigrad]

$$\cos 2x = -\frac{23}{25}$$

$$\tan 2x = \frac{\sin 2x}{\cos 2x}$$

All the rest look good.

 

[DethRose]

why are those 2 like that?

 

[DethRose]

how can cosx and cos2x both be 1/5?

Also here's how i did those 2 questions I am just wondering why they are wrong?

cos2x=1-2sin^2x
=1-2(-2 squareroot 6/5)squared
=1-2(24/25)
=23/25

tan2x=2tanx/1-tan^2x
=2(-2 squareroot 6)/1-6
=-4 squareroot 6/-5
=4 squareroot 6/5

 

[courtrigrad]

cos2x=1-2sin^2x
=1-2(-2 squareroot 6/5)squared
=1-2(24/25)
=23/25

I said that it should be a -23/25.

And you are right.

 

[Aviig]

To quote courtigrad:
"$$\sin 2x = 2\sin x \cos x$$"


I think this might be a problem for some, because I was doing problems similar to this the other day, and only today did we begin learning the double-angle, half-angle, and power down formulas.