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C2 maths help

Why does (sin 2x) / (cos 2x) = tan 2x . I thought the answer was tan x because the 2s cancel each other out
Original post by man111111
Why does (sin 2x) / (cos 2x) = tan 2x . I thought the answer was tan x because the 2s cancel each other out


nah, g math don't work like that.

sin() \sin () is a function.

cos() \cos() is a function

You are applying 2x 2x to that function, you can't just cancel the (2x). Imagine the trig functions as vacuum cleaners, once you suck something up with the vacuum cleaner you can't get it back in it's purest form.
(edited 6 years ago)
Reply 2
if it was 2sin x and 2cos x you could've cancel the 2s out, ah man i do miss a level maths :')
As a more general formula

sin()cos()=tan() \frac{\sin()}{\cos()} = \tan()

So....

sin(1)cos(1)=tan(1) \frac{\sin(1)}{\cos(1)} = \tan(1)

sin(x)cos(x)=tan(x) \frac{\sin(x)}{\cos(x)} = \tan(x)

sin(2x)cos(2x)=tan(2x) \frac{\sin(2x)}{\cos(2x)} = \tan(2x)

sin(3x)cos(3x)=tan(3x) \frac{\sin(3x)}{\cos(3x)}= \tan(3x)

In words, Sine of anything, divided by cosine of that same anything equals tangent of that same anything

Have I left anything out? :wink:

nah that's moist :s-smilie:
(edited 6 years ago)
Original post by Apachai Hopachai
As a more general formula

sin()cos()=tan() \frac{\sin()}{\cos()} = \tan()

So....

sin(1)cos(1)=tan(1) \frac{\sin(1)}{\cos(1)} = \tan(1)

sin(x)cos(x)=tan(x) \frac{\sin(x)}{\cos(x)} = \tan(x)

sin(2x)cos(2x)=tan(2x) \frac{\sin(2x)}{\cos(2x)} = \tan(2x)

sin(3x)cos(3x)=tan(3x) \frac{\sin(3x)}{\cos(3x)}= \tan(3x)

In words, Sine of anything, divided by cosine of that same anything equals tangent of that same anything

Have I left anything out? :wink:

nah that's moist :s-smilie:


That "anything" cannot equal π2+πn\frac{\pi}{2}+\pi n for nZn\in \mathbb{Z} otherwise you get division by 0
Original post by RDKGames
That "anything" cannot equal π2+πn\frac{\pi}{2}+\pi n for nZn\in \mathbb{Z} otherwise you get division by 0


yes
Original post by RDKGames
That "anything" cannot equal π2+πn\frac{\pi}{2}+\pi n for nZn\in \mathbb{Z} otherwise you get division by 0


you had to confuse it..

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