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chels
Group Title
provide that this equation is an identity.
sin(x)cos(x) / tan(x) = 1  sin^2(x)
 3 years ago
 3 years ago
chels Group Title
provide that this equation is an identity. sin(x)cos(x) / tan(x) = 1  sin^2(x)
 3 years ago
 3 years ago

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lokisan Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{\sin x \cos x}{\tan x } =\frac{\sin x \cos x}{\frac{\sin x }{\cos x}}=\frac{\cos x}{\frac{1}{\cos x}}=\cos^2 x = 1\sin^2 x\]
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
i can't scroll..
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Try refreshing the page, or log out and log back in.
 3 years ago

beens Group TitleBest ResponseYou've already chosen the best response.0
sinx cosx/tanx=sinx cosx cosx/sinx =cosxcosx =cos^2x =1sin^x
 3 years ago

beens Group TitleBest ResponseYou've already chosen the best response.0
sorry...1sin^2x
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
how does sinxcosx/tanx = sinx?
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
It doesn't. It equals cos^2x
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
just kidding, i understand how you wrote it now
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
I like how you stalked me then...lol
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
haha sorry. i got confused.
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
1 / sec(x)  1 minus 1 / sec(x) + 1 = 2cot^2(x)
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Can you type out the question using the equation editor?
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
\[1\div \sec(x1)  1 \div \sec(x+1) = 2\cot^2x\]
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
in the denominator it's supposed to be sec(x)  1 and sec(x) + 1
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Yeah...I was thinking that.
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Okay, it's easy now...
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{1}{\sec x 1}\frac{1}{\sec x +1}=\frac{\sec x +1  (\sec x 1)}{\sec^2 x1}=\frac{2}{\tan^2 x}\]
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
\[=2\cot^2 x\]
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Afte\[1+\tan^2 x=\sec^2 x \rightarrow 1\sec^2x = \tan^2 x\]r putting it over a common denominator, I use the fact that
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Then cot^2(x) is just from the definition, \[\cot x = \frac{1}{\tan x}\]
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Once you're done, let me know if it works for you or not.
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
okay, i understand that one
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
\[\cos(\pi/2  x)\cos(x) = \sin(2x) \div 2\]
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Okay, you have to identify a few things here. (1)\[\cos(\frac{\pi}{2}x)=\sin x\](2)\[\cos(x)=\cos x\]Then the lefthand side becomes,\[\cos(\frac{\pi}{2}x)\cos(x)=\sin x \cos x \]If you multiply both sides by 2, you get\[2\sin x \cos x = \sin 2x\]which is true (since this is the expansion of sin(2x)).
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
how did you identify (1)?
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
If you want to see why cos(pi/2  x) = sin(x), draw coordinate axes, the unit circle, and swing the radial arm from 0 to pi (180 degrees) and then reverse a bit (that bit is equal to x). The angle in the triangle ...I'm doing that now...
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
is x. I'll draw a pic.
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Actually, instead of a geometrical explanation (which will take forever online) do you know your double angle formulae?
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
yeah, i get it now. i just had to go through it all
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
\[\tan(x \div 2) \cos^2(x)  \tan(x \div 2) = \sin(x) \div \sec(x) minus \sin (x)\]
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
sin(x) is subtracted from sin(x) / sec(x) in the second equation.
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Okay, sorry, had to do something. Here we go...
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Recognize, on the lefthand side, you can factor as\[\tan^2\frac{x}{2}(\cos^2 x1)=\tan^2 x (\sin^2 x)\]
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Also, on the righthand side, \[\frac{\sin x }{\sec x}\sin x = \sin x (\cos x  1)\]We have then\[\tan^2 \frac{x}{2}(\sin^2 x)=\sin x (\cos x 1)\]
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
got it so far..
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Divide both sides by (sin^2(x)) to get\[\tan^2\frac{x}{2}=\frac{1\cos x}{\sin x}=\frac{1(\cos^2 \frac{x}{2} \sin ^2 \frac{x}{2})}{2\sin \frac{x}{2}\cos \frac{x}{2}}\]
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
But, in the numerator, 1cos^2 (x/2) = sin^2(x/2). So after expanding the numerator and doing this, we have\[\tan \frac{x}{2}=\frac{2\sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\]which is, tan(x/2) by definition.
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Forget the tan^2(x/2) thing I started with  transcribing notes to this site leads to screwups.
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
find exact value of (pi/12)
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Use the fact that 2 x pi/12 = pi/6 which is 30 degrees. We know sin(30 deg) = sin(30 deg) is. You can use the double angle formula to link the two.
 3 years ago

chels Group TitleBest ResponseYou've already chosen the best response.0
square root of 3 over 4?
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
Do you want to try that and see how you go. I need a shower.
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
The sine of 30 degrees is 1/2. Were you ever shown how to construct the basic angles from triangles? You can always work out 30, 45 and 60, and various combinations, using two triangles. One is a rightangled one with sides 1, 1 and hypotenuse sqrt(2). The angles are 90, 45 and 45. You can construct the 30 and 60 from taking an equilateral triangle of side length 2. Then cut it in half. You'll two rightangled triangles. The angles in each will be 30, 60 and 90. The sides will be 1, sqrt(3) and 2 (2 is the hypotenuse).
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
You can read off your sine, cosine and tangents from them.
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
chels, you can do this with the double angle formula for sine:\[\sin(\frac{\pi}{12})=\sin(\frac{\pi}{6}\frac{\pi}{4})\]\[=\sin \pi/6 \cos \pi/4  \cos \pi/6 \sin \pi/4\]\[=\frac{1}{2}\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}}=\frac{1\sqrt{3}}{2\sqrt{2}}=\frac{\sqrt{2}\sqrt{6}}{4}\]
 3 years ago

lokisan Group TitleBest ResponseYou've already chosen the best response.0
You could also use the combination,\[\frac{\pi}{4}\frac{\pi}{3}\]
 3 years ago
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