Trigonometric Functions-Tech Free question

[squance]

My sis is stuck on this question and would like some help please.

Show that:

2 sin^-1 (3/5) = sin^-1 (24/25)

[TrueTears]

Let

and

so

and

sub these in yields



so basically we have to prove

both sides yields



using sin compound angle formula yields



now (to work out cos(a) just draw the triangle and we know its a Pythagorean triple 3,4,5 )
so
(again same principle, draw the triangle and work it out, notice this is also a Pythagorean triple 7,24,25 )


subbing these values in to yields

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.

[squance]

when substituting to get 2a=b are you substituting 2(sinA)=sinB or what? Thanks

[TrueTears]

nope you're just subbing in because that is 'a'

the just comes from 'sin'ing both sides of and this is used later in the question (during the expansion of the compound angle formula)

maybe i shouldn't have put so early in the working haha

[squance]

Hehe. Thanks.

Another question needing help please:

Simplify

(sin3A/sinA) - (cos3A/cosA)

[TrueTears]

the crucial step here is the very first one, you have to notice sin3AcosA-sinAcos3A is the compound angle formula of sin(x-y)

[squance]

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

[TrueTears]

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

[TrueTears]

= = = =

now notice the numerator, let a = cosx



the denominator can be changed to , now notice this is DOPS(difference of perfect squares) which can be changed to

so the fraction now becomes as required.

[squance]

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!

[forthelolz]

FOR THE LOLZ:

[squance]

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?

[TrueTears]

hm i seem to get

I'll show my working now.

[TrueTears]

so

let

so now becomes sin(2a) (just subbing in a for )

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 (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 , so we know that it is , 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

Subbing this back in

[Over9000]

Yeh, I got same answer as TT, so im pretty sure its right