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

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

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

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

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

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

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

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

chels
 4 years ago
Best ResponseYou've already chosen the best response.0haha sorry. i got confused.

chels
 4 years ago
Best ResponseYou've already chosen the best response.01 / sec(x)  1 minus 1 / sec(x) + 1 = 2cot^2(x)

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

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

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

lokisan
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah...I was thinking that.

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

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

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

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

chels
 4 years ago
Best ResponseYou've already chosen the best response.0okay, i understand that one

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

lokisan
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, 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)).

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

lokisan
 4 years ago
Best ResponseYou've already chosen the best response.0If 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...

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

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

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

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

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

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

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

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

lokisan
 4 years ago
Best ResponseYou've already chosen the best response.0But, 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.

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

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

lokisan
 4 years ago
Best ResponseYou've already chosen the best response.0Use 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.

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

lokisan
 4 years ago
Best ResponseYou've already chosen the best response.0The 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).

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

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

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