Author Topic: Circular functions (Read 12446 times) Tweet Share

Flaming_Arrow · « **Reply #15 on:** March 02, 2009, 06:37:01 pm »

If $\ x= \sec \theta - \tan \theta$ , prove that $\ {x} + \frac {1}{x} = 2 sec \theta$ and find a simple expression for $\ {x} - \frac {1}{x}$

$\sec \theta - \tan \theta + \frac{1}{\sec \theta - \tan \theta}= 2 \sec \theta$

$\frac{(\sec \theta - \tan \theta)^2 + 1}{\sec \theta - \tan \theta}= 2 \sec \theta$

$\frac{\sec^2 \theta - 2\tan\sec\theta + \tan^2\theta + 1}{\sec \theta - \tan \theta}= 2 \sec \theta$

$\frac{\sec^2 \theta - 2\tan\sec\theta + sec^2 \theta}{\sec \theta - \tan \theta}= 2 \sec \theta$

$\frac{2\sec^2 \theta - 2\tan\sec\theta}{\sec \theta - \tan \theta}= 2 \sec \theta$

$\frac{2\sec \theta (\sec- \tan\theta)}{\sec \theta - \tan \theta}= 2 \sec \theta$

$2 \sec \theta= 2 \sec \theta$

LHS=RHS QED

d0minicz · « **Reply #16 on:** March 02, 2009, 06:43:50 pm »

thanks heaps but i dont get where the identity $\cos2x = 1- 2sin^2x$ comes from

Flaming_Arrow · « **Reply #17 on:** March 02, 2009, 06:44:09 pm »

TrueTears · « **Reply #18 on:** March 02, 2009, 06:46:14 pm »

its one of the identities you should know

remember: $cos2x = cos^2x - sin^2x = 1-2sin^2x = 2cos^2x-1$

TrueTears · « **Reply #19 on:** March 02, 2009, 06:49:35 pm »

You can also easily derive those by doing

$cos(2x) = cos(x+x) = cosxcosx - sinxsinx = cos^2x - sin^2x$

and for $2cos^2x-1$

$cos^2x - sin^2x = cos^2x - (1 - cos^2x) = 2cos^2x - 1$

and for $1-2sin^2x$

$cos^2x - sin^2x = (1 - sin^2x) - sin^2x = 1 - 2sin^2x$

d0minicz · « **Reply #20 on:** March 02, 2009, 06:52:05 pm »

thank you =]

do you have a list of identities that we need to know?

TrueTears · « **Reply #21 on:** March 02, 2009, 06:53:22 pm »

Sure, i'll put them all up, i made a summary sheet, give me a sec and type em XD

TrueTears · « **Reply #22 on:** March 02, 2009, 06:57:47 pm »

$cos(x-y) = cosxcosy + sinxsiny$

$cos(x+y) = cosxcosy - sinxsiny$

$sin(x-y) = sinxcosy - sinycosx$

$sin(x+y) = sinxcosy + sinycosx$

$cos(2x) = cos^2x - sin^2x = 1-2sin^2x = 2cos^2x-1$

$sin(2x) = 2sinxcosx$

$tan(x-y) = \frac{tanx - tany}{1+tanxtany}$

$tan(x+y) = \frac{tanx+tany}{1-tanxtany}$

$tan(2x) = \frac{2tanx}{1-tan^2x}$

$sin^2x + cos^2x = 1$

so basically those are your common sin and cos and tan ones

Flaming_Arrow · « **Reply #23 on:** March 02, 2009, 06:58:58 pm »

thanks truetears!

d0minicz · « **Reply #24 on:** March 02, 2009, 07:33:04 pm »

thanks =] !

any others we should familiarize ourselves with?

TrueTears · « **Reply #25 on:** March 02, 2009, 07:52:16 pm »

Well from $sin^2x + cos^2x = 1$ you can derive some others...

$cosec^2x = 1+ cot^2x$

and

$sec^2x = 1+tan^2x$

d0minicz · « **Reply #26 on:** March 02, 2009, 08:01:12 pm »

hey can you please show me how to find

$\ tan \frac {5\pi}{12}$ using compound angle formulas

TrueTears · « **Reply #27 on:** March 02, 2009, 08:04:41 pm »

sure,

so we know $\frac{5\pi}{12} = \frac{\pi}{6} + \frac{\pi}{4}$

Now, you can sub those in the tan(x+y) formula to get $tan(\frac{\pi}{6} + \frac{\pi}{4})$ right? Then use the formula $tan(x+y) = \frac{tanx+tany}{1-tanxtany}$
I think the rest you can do

Remember when doing these questions always think about $\frac{\pi}{4}, \frac{\pi}{6}, \frac{\pi}{3} and \frac{\pi}{2}$ , ie your exact values

d0minicz · « **Reply #28 on:** March 02, 2009, 08:58:28 pm »

If $\tan y = 2.4$ and $\ tan x = -0.75$

find $\tan (x+2y)$
need working out
thnks

TrueTears · « **Reply #29 on:** March 02, 2009, 09:19:13 pm »

k so here we just 'expand' tan(x+2y) right?

so $tan(x+2y) = \frac{tanx + tan2y}{1-tanxtan2y}$

However we are not given the value of $tan(2y)$ , but we are given the value of $tany$ , so using that information we can use the tan double angle formula.

$tan(2y) = \frac{2tany}{1-tan^2y}$

subbing in tany = 2.4 in yields $tan(2y) = -\frac{120}{119}$

then subbing this value (- $\frac{120}{119}$ ) back into the original we get:

$tan(x+2y) = \frac{-0.75 + -\frac{120}{119}}{1--0.75 \times -\frac{120}{119}}$

working that out equals = - $\frac{837}{116}$

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