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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: squance on March 14, 2009, 04:20:58 pm

Title: Trigonometric Functions-Tech Free question
Post by: squance on March 14, 2009, 04:20:58 pm: My sis is stuck on this question and would like some help please.

Show that:

2 sin^-1 (3/5) = sin^-1 (24/25)
Title: Re: Trigonometric Functions-Tech Free question
Post by: TrueTears on March 14, 2009, 04:27:26 pm: Let $sin^{-1}\frac{3}{5} = a$

and $sin^{-1}=\frac{24}{25} = b$

so $sina = \frac{3}{5}$

and $sinb = \frac{24}{25}$

sub these in yields $2a = b$

$2a - b = 0$

so basically we have to prove $2a - b = 0$

$sin$ both sides yields

$sin(2a-b) = sin0 = 0$

using sin compound angle formula yields

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

now $sin(2a) = 2sin(a)cos(a) =$ $2\frac{3}{5}\frac{4}{5}$ (to work out cos(a) just draw the triangle and we know its a Pythagorean triple 3,4,5 $cos = \frac{adj}{hyp} = \frac{4}{5}$ )
so $sin(2a) = \frac{24}{25}$
$cos(b) = \frac{7}{25}$ (again same principle, draw the triangle and work it out, notice this is also a Pythagorean triple 7,24,25 )
$cos(2a) = 1-2sin^2(a) = 1-2(\frac{3}{5})^2 = \frac{7}{25}$

subbing these values in to $sin(2a)cos(b) - sin(b)cos(2a)$ yields

$sin(2a-b) = \frac{24}{25} \times \frac{7}{25} - \frac{24}{25} \times \frac{7}{25} = 0$ as required.

Whenever I see these types of questions i always let the inverse equal say a or b and then try to get a compound angle formula or double angle formula going.
Title: Re: Trigonometric Functions-Tech Free question
Post by: squance on March 14, 2009, 05:33:31 pm: when substituting to get 2a=b are you substituting 2(sinA)=sinB or what? Thanks :)
Title: Re: Trigonometric Functions-Tech Free question
Post by: TrueTears on March 14, 2009, 05:36:21 pm: nope you're just subbing in $sin^{-1}\frac{3}{5}$ because that is 'a'

the $sina = \frac{3}{5}$ just comes from 'sin'ing both sides of $sin^{-1}\frac{3}{5} = a$ and this is used later in the question (during the expansion of the compound angle formula)

maybe i shouldn't have put $sina = \frac{3}{5}$ so early in the working haha
Title: Re: Trigonometric Functions-Tech Free question
Post by: squance on March 14, 2009, 05:56:38 pm: Hehe. Thanks.

Another question needing help please:

Simplify

(sin3A/sinA) - (cos3A/cosA)
Title: Re: Trigonometric Functions-Tech Free question
Post by: TrueTears on March 14, 2009, 06:09:21 pm: $= \frac{sin3AcosA-sinAcos3A}{sinAcosA} = \frac{sin(3A-A)}{sinAcosA} = \frac{sin2A}{sinAcosA} = \frac{2sinAcosA}{sinAcosA} = 2$

the crucial step here is the very first one, you have to notice sin3AcosA-sinAcos3A is the compound angle formula of sin(x-y) :P
Title: Re: Trigonometric Functions-Tech Free question
Post by: squance on March 14, 2009, 06:19:22 pm: Thanks once again.
My sister once again appreciates your help.

Last question she has; Promise.

Prove the identity:

(cot x + cosec x)^2 = 1 + cos x / 1 - cos x
Title: Re: Trigonometric Functions-Tech Free question
Post by: TrueTears on March 14, 2009, 06:22:30 pm: Quote from: squance on March 14, 2009, 06:19:22 pm
Thanks once again.
My sister once again appreciates your help.

Last question she has; Promise.

Prove the identity:

(cot x + cosec x)^2 = 1 + cos x / 1 - cos x
Lol nps, ask as much as you want, these questions are fun :)
Title: Re: Trigonometric Functions-Tech Free question
Post by: TrueTears on March 14, 2009, 06:29:37 pm: $(cotx + cosecx)^2$ = $(\frac{cosx}{sinx} + \frac{1}{sinx})^2$ = $(\frac{cosx+1}{sinx})^2$ = $\frac{(cosx+1)^2}{(sinx)^2}$ = $\frac{cos^2x + 2cosx +1}{sin^2x}$

now notice the numerator, let a = cosx

$a^2+2a+1 = (a+1)^2 = (cosx+1)^2$

the denominator can be changed to $1-cos^2x$ , now notice this is DOPS(difference of perfect squares) which can be changed to $(1+cosx)(1-cosx)$

so the fraction now becomes $\frac{(cosx+1)^2}{(1+cosx)(1-cosx)} = \frac{1+cosx}{1-cosx}$ as required.
Title: Re: Trigonometric Functions-Tech Free question
Post by: squance on March 14, 2009, 06:47:42 pm: YAY!
Many thanks TrueTears!

My sister actually got up to your second line of working, but then she forgot to sub in the cos x back into the a.

LOL

THANK YOU!
YOUR AWESOME!
Title: Re: Trigonometric Functions-Tech Free question
Post by: forthelolz on March 15, 2009, 02:18:50 pm: FOR THE LOLZ:

(http://img10.imageshack.us/img10/8294/fortehlolz.jpg)
Title: Re: Trigonometric Functions-Tech Free question
Post by: squance on March 15, 2009, 08:55:35 pm: Hey.
My sister has another question she would like checked.

Evaluate:
sin(2 cos^-1 (-5/13))

Her answer was sin x = (sqrt 651)/26

Can anyone please verfiy this?
Title: Re: Trigonometric Functions-Tech Free question
Post by: TrueTears on March 15, 2009, 09:03:28 pm: hm i seem to get $\frac{-120}{169}$

I'll show my working now.
Title: Re: Trigonometric Functions-Tech Free question
Post by: TrueTears on March 15, 2009, 09:10:11 pm: so $sin(2cos^{-1}\frac{-5}{13})$

let $a = cos^{-1}\frac{-5}{13} \implies cosa = \frac{-5}{13}$

so $sin(2cos^{-1}\frac{-5}{13})$ now becomes sin(2a) (just subbing in a for $cos^{-1}\frac{-5}{13}$ )

now use the double angle formula for sin(2a)

this yields: sin(2a) = 2sinacosa

we have to work out sina

again draw the triangle and you can see $sina = \frac{12}{13}$ (again Pythagorean triple 5,12,13)

now be careful here, we need to see if sina is negative or positive. so to work that out we need to know the range of $cos^{-1}$ , so we know that it is $[0,\pi]$ , now this means that the values of cos(a) are in the 1st and 2nd quadrants, however cos(a) is a negative value which means it must be in the 2nd quadrant, therefore sin(a) is also in the 2nd quadrant and here it is positive. So we just take $+\frac{12}{13}$

Subbing this back in $sin(2a) = 2 \times \frac{12}{13} \times \frac{-5}{13} = \frac{-120}{169}$
Title: Re: Trigonometric Functions-Tech Free question
Post by: Over9000 on March 15, 2009, 09:11:49 pm: Yeh, I got same answer as TT, so im pretty sure its right
Title: Re: Trigonometric Functions-Tech Free question
Post by: squance on March 15, 2009, 09:19:49 pm: Cool. Awesome.

My sister said that for these questions, they didn't typically need the double angle..I guess she didn't know what to do when the 2 was in front of the cos.
Thanks TrueTears and Over9000
Title: Re: Trigonometric Functions-Tech Free question
Post by: TrueTears on March 15, 2009, 09:25:02 pm: Yeah hehe, it's good to know those formulas, comes in handy :)
Title: Re: Trigonometric Functions-Tech Free question
Post by: forthelolz on March 15, 2009, 09:26:03 pm: FOR THE LOLZ

(http://img15.imageshack.us/img15/293/forthelolz.jpg)
Title: Re: Trigonometric Functions-Tech Free question
Post by: Over9000 on March 15, 2009, 09:27:18 pm: Just make sure that when you have something like $sin(2cos^{-1}(x))$ you always apply double angle formula, and dont include the 2 in your working out for x until the end, this was a mistake I made, which TT helped me with.