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