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Rosy95
PreCal Help Please..I am so stuck on what to do
Use the equation below to find the value of sin x. \[\frac{ \tan x }{ \cot x } - \frac{ \sec x }{ \cos x }= \frac{ 2 }{ \csc x }\]
write everything in terms of sin and cos
That is what I am not understanding. I have a really hard time writting out what it is supposed to be once you start changing everything.
Hint: Write it this way. Then simply change everything to sin and cos \[\tan(x) \div \cot(x) - \sec(x) \div \cos(x) = 2 \div \csc(x)\]
So how would you change it?
\[\frac{\sin(x)}{\cos(x)} \div \frac{\cos(x)}{\sin(x)} - \frac{1}{\cos(x)} \div \cos(x) = 2 \div \frac{1}{\sin(x)}\]
so could the first two fracts cancel out?
How do you divide: \[\frac{1}{4} \div \frac{7}{6}\]
Why not? BTW i missed the lessons over this stuff so i am really lost on almost all of it.
Answer the previous question. How do you divide that?
you switch the #'s of the second fraction so it is now 6/7 and you change divide to multiply. you now have 1/4 X 6/7 which is 6/28 simplified to 3/14
Yes, so do that same thing with the trig equation
so it is now \[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos x }\div \cos x=2\div \frac{ 1 }{ \sin x }\]??
Yes, but do the rest of them. Change all of the divisions into multiplication.
so it looks like\[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ cosx }\times \frac{ cosx }{ 1 }=2\times \frac{ sinx }{ 1 }\]
If so, wouldn't 1/cosx times cosx/1 cancel out?
no, you did the cos(x)/1 incorrectly. cos(x) is already equal to cos(x)/1. You have to flip it.
oh duh ok so then it is \[\frac{ \sin^2 x}{ \cos^2x } -\frac{ 1 }{ \cos x }\times \frac{ 1 }{ \cos x }= 2\times \frac{ sinx }{ 1 }\]
so do the mult fractions combine to get -1/cos^2x
Yes, so what do you have after that?
\[\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos^2x }= 2\times \frac{ sinx }{ 1 }\]
Okay, do you know how to combine fractions on the left side?
couldn't you turn 2x(sinx/1) into just 2sinx? and Im not sure what to do on the other side
On the left side, you have to combine fractions if they have the same denominator.
For example: \[\frac{1}{4} + \frac{x}{4} = \frac{1 + x}{4}\]
The general rule for combining fractions with the same denominator is: \[\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}\]
so it is now \[\frac{ \sin^2x-1 }{ \cos^2x }=2sinx\]
or is the 2sin x supposed to be sin^2x
Now change the 1 to \(\sin^2x + \cos^2x\)
No, you have it right. 2sin(x)
wait, you put sin^2x + cos^2x instead of one? what is the new equation?
Yes, you put \(\sin^2x + \cos^2x\) in place of 1 because they are equal.
What do you get afterwards?
\[\frac{ \sin^2x-\sin^2x+\cos^2x }{ \cos^2x }=2sinx\] then you can subtract the sin^2x to get \[\frac{ \cos^2x }{ \cos^2x }=2sinx\] then if thats the case you would just have 2sinx
Actually, if you do it correctly you get: \[-\frac{ \cos^2x }{ \cos^2x }=2\sin x\]
Which simplifies to \[-1 = 2 \sin x\] Remember that we are still solving for x
You were supposed to get this after substituting the 1: \[\frac{ \sin^2x-(\sin^2x+\cos^2x) }{ \cos^2x }=2\sin x\] Then get : \[\frac{ \sin^2x-\sin^2x-cos^2x }{ \cos^2x }=2\sin x\]
or no sinx would be -1/2
Ultimately you end up isolating the sin(x) on the right side: \[-\frac{1}{2} = \sin x\]
Now all you have to do is take the inverse sin of both sides to get \[\sin^{-1}\left(-\frac{1}{2}\right) = x\]
So evaluate the left side using your calc. Let me know what you get. Also make sure it is in radian mode.
And thank you for your help :)
Well, in exact mode, you should have gotten \[x = -\frac{\pi}{6}\]
Nevertheless:\[-\frac{\pi}{6} \approx -.52\]