Trig identities *eek*

**daunder** · April 25th 2009, 12:41 PM

A) $(cscx -cotx)^2 = \frac{1-cosx}{1+cosx}$

B) $sin2x = \frac{2tanx}{sec^2x}$

C) $cos(x+y) cos(x-y) = cos^2x + cos^2y -1$

**VonNemo19** · April 25th 2009, 01:43 PM

In regards to c)

(3)

So therefore,

(

)(

)=

(cos^2a)(cos^2b)-(sin^2a)(sin^2b)=

(cos^2a)(cos^2b)-(1-cos^2a)(1-cos^2b)=

(cos^2a)(cos^2b)-1+cos^2a+cos^2b-(cos^2a)(cos^2b)=

cos^2a+cos^2b-1

**Soroban** · April 25th 2009, 02:30 PM

Hello, daunder!

$A)\;\;(\csc x -\cot x)^2 \:=\: \frac{1-\cos x}{1+\cos x}$

The left side is: . $\left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right)^2 \;=\;\left(\frac{1-\cos x}{\sin x}\right)^2 \;=\;\frac{(1-\cos x)^2}{\sin^2\!x}$

. . $= \;\frac{(1-\cos x)^2}{1-\cos^2\!x} \;=\;\frac{(1-\cos x)(1 - \cos x)}{(1-\cos x)(1 + \cos x)} \;=\;\frac{1-\cos x}{1 + \cos x}$

$B)\;\;\sin2x \:=\:\frac{2\tan x}{\sec^2\!x}$

The right side is: . $\frac{2\,\dfrac{\sin x}{\cos x}}{\dfrac{1}{\cos^2\!x}} \;=\;2\sin x\cos x \;=\;\sin2x$

$C)\;\;\cos(x+y) \cos(x-y) \:=\: \cos^2\!x + \cos^2\!y -1$

This requires a Compound-Angle Identity: . $\cos(A \pm B) \;=\;\cos A\cos B \mp \sin A\sin B$

The left side is: . $(\cos x\cos y - \sin x\sin y)(\cos x\cos y + \sin x\sin y)$

Multiply: . $(\cos x\cos y )^2 - (\sin x\sin y)^2$

. . . . . $=\;\cos^2\!x\cos^2\!y - \sin^2\!x\sin^2\!y$

. . . . . $= \;\cos^2\!x\cos^2\!y - (1-\cos^2\!x)(1 - \cos^2\!y)$

. . . . . $=\;\cos^2\!x\cos^2\!y - (1 - \cos^2\!x - \cos^2\!y + \cos^2\!x\cos^2\!y)$

. . . . . $=\;\cos^2\!x\cos^2\!y - 1 + \cos^2\!x + \cos^2\!y - \cos^2\!x\cos^2\!y$

. . . . . $=\; \cos^2\!x + \cos^2\!y - 1$

**VonNemo19** · April 25th 2009, 02:40 PM

This is unconventional I'm sure but in regars to a):

(cscx-cotx)^2=csc^2x-2cscxcotx+cot^2x=

(1/sin^2x)-2(cosx/sin^2x)+(cos^2x/sin^2x)=

(1-2cosx+cos^2x)/sin^2x=(1-cosx)/(1+cosx)

multiplying both sides by 1+cosx yields

(1-cosx-cos^2x+cos^3x)/(1-cos^2x)=1-cosx

(remember that sin^2x=1-cos^2x)

Now Multiply both sides by 1-cos^2x

1-cosx-cos^2x+cos^3x=1-cosx-cos^2x+cos^3x

I know that this is probably a strange way to prove this identity, but I learned everything I know about math from a prison library.

**Soroban** · April 25th 2009, 04:45 PM

Hello, VonNemo19!

This is unconventional I'm sure, but in regards to a):

$(\csc x-\cot x)^2\:=\: \csc^2\!x-2\csc x\cot x+\cot^2\!x$

. . $=\;\left(\frac{1}{\sin^2\!x}\right) - 2\left(\frac{\cos x}{\sin^2\!x}\right) + \left(\frac{\cos^2\!x}{\sin^2\!x}\right)$

. . $=\;\frac{1-2\cos x+\cos^2\!x}{\sin^2\!x}$

multiplying both sides by $1+\cos x$ . . . . no

The accepted rule is that we never play with both side of the "equation".
It is not an equation because it has not been established that the two sides are equal.
In fact, that is what we are asked to prove. .**

You could continue like this: . $\frac{(1-\cos x)^2}{1-\cos^2x} \;=\;\frac{(1-\cos x)(1-\cos x)}{(1-\cos x)(1 + \cos x)} \;=\;\frac{1-\cos x}{1 + \cos x}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Imagine there is philosophy test queston: Prove that there is a God.

And I answer: There is a God, because in the Bible, it says:
. . . . . . . . . . "In the Beginning, God created Heaven and Earth."
. . . . . . . . . . And God would never lie.

**daunder** · April 25th 2009, 06:33 PM

Thanks Soroban

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