chels
  • chels
provide that this equation is an identity. sin(x)cos(x) / tan(x) = 1 - sin^2(x)
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{\sin x \cos x}{\tan x } =\frac{\sin x \cos x}{\frac{\sin x }{\cos x}}=\frac{\cos x}{\frac{1}{\cos x}}=\cos^2 x = 1-\sin^2 x\]
chels
  • chels
i can't scroll..
anonymous
  • anonymous
Try refreshing the page, or log out and log back in.

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anonymous
  • anonymous
sinx cosx/tanx=sinx cosx cosx/sinx =cosxcosx =cos^2x =1-sin^x
anonymous
  • anonymous
sorry...1-sin^2x
chels
  • chels
how does sinxcosx/tanx = sinx?
anonymous
  • anonymous
It doesn't. It equals cos^2x
chels
  • chels
just kidding, i understand how you wrote it now
anonymous
  • anonymous
I like how you stalked me then...lol
chels
  • chels
haha sorry. i got confused.
chels
  • chels
1 / sec(x) - 1 minus 1 / sec(x) + 1 = 2cot^2(x)
anonymous
  • anonymous
Can you type out the question using the equation editor?
chels
  • chels
\[1\div \sec(x-1) - 1 \div \sec(x+1) = 2\cot^2x\]
anonymous
  • anonymous
okay
chels
  • chels
in the denominator it's supposed to be sec(x) - 1 and sec(x) + 1
anonymous
  • anonymous
Yeah...I was thinking that.
chels
  • chels
yeahh.
anonymous
  • anonymous
Okay, it's easy now...
anonymous
  • anonymous
\[\frac{1}{\sec x -1}-\frac{1}{\sec x +1}=\frac{\sec x +1 - (\sec x -1)}{\sec^2 x-1}=\frac{2}{\tan^2 x}\]
anonymous
  • anonymous
\[=2\cot^2 x\]
anonymous
  • anonymous
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
anonymous
  • anonymous
Then cot^2(x) is just from the definition, \[\cot x = \frac{1}{\tan x}\]
anonymous
  • anonymous
Once you're done, let me know if it works for you or not.
chels
  • chels
okay, i understand that one
chels
  • chels
\[\cos(\pi/2 - x)\cos(-x) = \sin(2x) \div 2\]
anonymous
  • anonymous
Okay, you have to identify a few things here. (1)\[\cos(\frac{\pi}{2}-x)=\sin x\](2)\[\cos(-x)=\cos x\]Then the left-hand 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
  • chels
how did you identify (1)?
anonymous
  • anonymous
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...
anonymous
  • anonymous
is x. I'll draw a pic.
anonymous
  • anonymous
Actually, instead of a geometrical explanation (which will take forever online) do you know your double angle formulae?
chels
  • chels
yeah, i get it now. i just had to go through it all
anonymous
  • anonymous
Okay
chels
  • chels
\[\tan(x \div 2) \cos^2(x) - \tan(x \div 2) = \sin(x) \div \sec(x) minus \sin (x)\]
chels
  • chels
sin(x) is subtracted from sin(x) / sec(x) in the second equation.
anonymous
  • anonymous
Okay, sorry, had to do something. Here we go...
anonymous
  • anonymous
Recognize, on the left-hand side, you can factor as\[\tan^2\frac{x}{2}(\cos^2 x-1)=\tan^2 x (-\sin^2 x)\]
chels
  • chels
ok, yep.
anonymous
  • anonymous
Also, on the right-hand side, \[\frac{\sin x }{\sec x}-\sin x = \sin x (\cos x - 1)\]We have then\[\tan^2 \frac{x}{2}(-\sin^2 x)=\sin x (\cos x -1)\]
chels
  • chels
got it so far..
anonymous
  • anonymous
Divide both sides by (-sin^2(x)) to get\[\tan^2\frac{x}{2}=\frac{1-\cos x}{\sin x}=\frac{1-(\cos^2 \frac{x}{2}- \sin ^2 \frac{x}{2})}{2\sin \frac{x}{2}\cos \frac{x}{2}}\]
anonymous
  • anonymous
But, in the numerator, 1-cos^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.
anonymous
  • anonymous
Forget the tan^2(x/2) thing I started with - transcribing notes to this site leads to screw-ups.
chels
  • chels
ok
chels
  • chels
find exact value of (-pi/12)
anonymous
  • anonymous
function?
chels
  • chels
sin(-pi/12)
chels
  • chels
sorry
anonymous
  • anonymous
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.
chels
  • chels
square root of 3 over 4?
anonymous
  • anonymous
Do you want to try that and see how you go. I need a shower.
anonymous
  • anonymous
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 right-angled 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 right-angled triangles. The angles in each will be 30, 60 and 90. The sides will be 1, sqrt(3) and 2 (2 is the hypotenuse).
anonymous
  • anonymous
You can read off your sine, cosine and tangents from them.
anonymous
  • anonymous
chels, you can do this with the double angle formula for sine:\[\sin(-\frac{\pi}{12})=\sin(\frac{\pi}{6}-\frac{\pi}{4})\]\[=\sin \pi/6 \cos \pi/4 - \cos \pi/6 \sin \pi/4\]\[=\frac{1}{2}\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}}=\frac{1-\sqrt{3}}{2\sqrt{2}}=\frac{\sqrt{2}-\sqrt{6}}{4}\]
anonymous
  • anonymous
You could also use the combination,\[\frac{\pi}{4}-\frac{\pi}{3}\]

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