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Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
Use the equation below to find the value of sin x. \[\frac{ \tan x }{ \cot x }  \frac{ \sec x }{ \cos x }= \frac{ 2 }{ \csc x }\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
write everything in terms of sin and cos
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Hint: Write it this way. Then simply change everything to sin and cos \[\tan(x) \div \cot(x)  \sec(x) \div \cos(x) = 2 \div \csc(x)\]
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
So how would you change it?
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
\[\frac{\sin(x)}{\cos(x)} \div \frac{\cos(x)}{\sin(x)}  \frac{1}{\cos(x)} \div \cos(x) = 2 \div \frac{1}{\sin(x)}\]
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
so could the first two fracts cancel out?
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
How do you divide: \[\frac{1}{4} \div \frac{7}{6}\]
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
Why not? BTW i missed the lessons over this stuff so i am really lost on almost all of it.
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Answer the previous question. How do you divide that?
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
Oh wait I remember
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Yes, so do that same thing with the trig equation
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
so it is now \[\frac{ \sin^2x }{ \cos^2x }\frac{ 1 }{ \cos x }\div \cos x=2\div \frac{ 1 }{ \sin x }\]??
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Yes, but do the rest of them. Change all of the divisions into multiplication.
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
so it looks like\[\frac{ \sin^2x }{ \cos^2x }\frac{ 1 }{ cosx }\times \frac{ cosx }{ 1 }=2\times \frac{ sinx }{ 1 }\]
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
Did I do that right?
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
If so, wouldn't 1/cosx times cosx/1 cancel out?
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
no, you did the cos(x)/1 incorrectly. cos(x) is already equal to cos(x)/1. You have to flip it.
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
oh duh ok so then it is \[\frac{ \sin^2 x}{ \cos^2x } \frac{ 1 }{ \cos x }\times \frac{ 1 }{ \cos x }= 2\times \frac{ sinx }{ 1 }\]
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
so do the mult fractions combine to get 1/cos^2x
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
so then what do you do?
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Yes, so what do you have after that?
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ \sin^2x }{ \cos^2x }\frac{ 1 }{ \cos^2x }= 2\times \frac{ sinx }{ 1 }\]
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Okay, do you know how to combine fractions on the left side?
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
couldn't you turn 2x(sinx/1) into just 2sinx? and Im not sure what to do on the other side
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
What do I do next?
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
On the left side, you have to combine fractions if they have the same denominator.
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
For example: \[\frac{1}{4} + \frac{x}{4} = \frac{1 + x}{4}\]
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
The general rule for combining fractions with the same denominator is: \[\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}\]
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
so it is now \[\frac{ \sin^2x1 }{ \cos^2x }=2sinx\]
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
or is the 2sin x supposed to be sin^2x
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Now change the 1 to \(\sin^2x + \cos^2x\)
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
No, you have it right. 2sin(x)
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
wait, you put sin^2x + cos^2x instead of one? what is the new equation?
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Yes, you put \(\sin^2x + \cos^2x\) in place of 1 because they are equal.
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
What do you get afterwards?
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ \sin^2x\sin^2x+\cos^2x }{ \cos^2x }=2sinx\] 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
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Actually, if you do it correctly you get: \[\frac{ \cos^2x }{ \cos^2x }=2\sin x\]
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Which simplifies to \[1 = 2 \sin x\] Remember that we are still solving for x
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
so sinx is 2
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
You were supposed to get this after substituting the 1: \[\frac{ \sin^2x(\sin^2x+\cos^2x) }{ \cos^2x }=2\sin x\] Then get : \[\frac{ \sin^2x\sin^2xcos^2x }{ \cos^2x }=2\sin x\]
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
or no sinx would be 1/2
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Ultimately you end up isolating the sin(x) on the right side: \[\frac{1}{2} = \sin x\]
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Now all you have to do is take the inverse sin of both sides to get \[\sin^{1}\left(\frac{1}{2}\right) = x\]
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
So evaluate the left side using your calc. Let me know what you get. Also make sure it is in radian mode.
 one year ago

Rosy95 Group TitleBest ResponseYou've already chosen the best response.1
And thank you for your help :)
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Well, in exact mode, you should have gotten \[x = \frac{\pi}{6}\]
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.2
Nevertheless:\[\frac{\pi}{6} \approx .52\]
 one year ago
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