Question

provide that this equation is an identity.
sin(x)cos(x) / tan(x) = 1 - sin^2(x)

Answer 1

\[\frac{\sin x \cos x}{\tan x } =\frac{\sin x \cos x}{\frac{\sin x }{\cos x}}=\frac{\cos x}{\frac{1}{\cos x}}=\cos^2 x = 1-\sin^2 x\]

Answer 2

i can't scroll..

Answer 3

Try refreshing the page, or log out and log back in.

Answer 4

sinx cosx/tanx=sinx cosx cosx/sinx
=cosxcosx
=cos^2x
=1-sin^x

Answer 5

sorry...1-sin^2x

Answer 6

how does sinxcosx/tanx = sinx?

Answer 7

It doesn't. It equals cos^2x

Answer 8

just kidding, i understand how you wrote it now

Answer 9

I like how you stalked me then...lol

Answer 10

haha sorry. i got confused.

Answer 11

1 / sec(x) - 1 minus 1 / sec(x) + 1 = 2cot^2(x)

Answer 12

Can you type out the question using the equation editor?

Answer 13

\[1\div \sec(x-1) - 1 \div \sec(x+1) = 2\cot^2x\]

Answer 14

okay

Answer 15

in the denominator it's supposed to be sec(x) - 1 and sec(x) + 1

Answer 16

Yeah...I was thinking that.

Answer 17

yeahh.

Answer 18

Okay, it's easy now...

Answer 19

\[\frac{1}{\sec x -1}-\frac{1}{\sec x +1}=\frac{\sec x +1 - (\sec x -1)}{\sec^2 x-1}=\frac{2}{\tan^2 x}\]

Answer 20

\[=2\cot^2 x\]

Answer 21

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

Answer 22

Then cot^2(x) is just from the definition, \[\cot x = \frac{1}{\tan x}\]

Answer 23

Once you're done, let me know if it works for you or not.

Answer 24

okay, i understand that one

Answer 25

\[\cos(\pi/2 - x)\cos(-x) = \sin(2x) \div 2\]

Answer 26

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)).

Answer 27

how did you identify (1)?

Answer 28

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...

Answer 29

is x. I'll draw a pic.

Answer 30

Actually, instead of a geometrical explanation (which will take forever online) do you know your double angle formulae?

Answer 31

yeah, i get it now. i just had to go through it all

Answer 32

Okay

Answer 33

\[\tan(x \div 2) \cos^2(x) - \tan(x \div 2) = \sin(x) \div \sec(x) minus \sin (x)\]

Answer 34

sin(x) is subtracted from sin(x) / sec(x) in the second equation.

Answer 35

Okay, sorry, had to do something. Here we go...

Answer 36

Recognize, on the left-hand side, you can factor as\[\tan^2\frac{x}{2}(\cos^2 x-1)=\tan^2 x (-\sin^2 x)\]

Answer 37

ok, yep.

Answer 38

Also, on the right-hand side, \[\frac{\sin x }{\sec x}-\sin x = \sin x (\cos x - 1)\]We have then\[\tan^2 \frac{x}{2}(-\sin^2 x)=\sin x (\cos x -1)\]

Answer 39

got it so far..

Answer 40

Divide both sides by (-sin^2(x)) to get\[\tan^2\frac{x}{2}=\frac{1-\cos x}{\sin x}=\frac{1-(\cos^2 \frac{x}{2}- \sin ^2 \frac{x}{2})}{2\sin \frac{x}{2}\cos \frac{x}{2}}\]

Answer 41

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.

Answer 42

Forget the tan^2(x/2) thing I started with - transcribing notes to this site leads to screw-ups.

Answer 43

ok

Answer 44

find exact value of (-pi/12)

Answer 45

function?

Answer 46

sin(-pi/12)

Answer 47

sorry

Answer 48

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.

Answer 49

square root of 3 over 4?

Answer 50

Do you want to try that and see how you go. I need a shower.

Answer 51

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).

Answer 52

You can read off your sine, cosine and tangents from them.

Answer 53

chels, you can do this with the double angle formula for sine:\[\sin(-\frac{\pi}{12})=\sin(\frac{\pi}{6}-\frac{\pi}{4})\]\[=\sin \pi/6 \cos \pi/4 - \cos \pi/6 \sin \pi/4\]\[=\frac{1}{2}\frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}}=\frac{1-\sqrt{3}}{2\sqrt{2}}=\frac{\sqrt{2}-\sqrt{6}}{4}\]

Answer 54

You could also use the combination,\[\frac{\pi}{4}-\frac{\pi}{3}\]