# Trig IdentitiesWatch

This discussion is closed.
#1
Hey people!
Can you help with this question?

Show that (4 cos 2x)/(sin^2 2x) is identical to (cosec^2 x) - (sec^2 x)

It's really doing my head in!
0
14 years ago
#2
First expand sin2x and cos2x, then simplify.
0
14 years ago
#3
cos(2x) = cos^2(x) - sin^2(x)
sin(2x) = 2sin(x)cos(x)

4cos(2x)/sin^2(2x)
= 4[cos^2(x) - sin^2(x)] / [4sin^2(x)cos^2(x)]
= [cos^2(x) - sin^2(x)] / [sin^2(x)cos^2(x)]
= cosec^2(x) - sec^2(x)
0
#4
thank you people!!
0
14 years ago
#5
0
#6
(Original post by mvmv)
sorry, but what?
0
14 years ago
#7
How is it shown that: 'cos(2x) ≡ cos^2(x) - sin^2(x)' ?
0
14 years ago
#8
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
0
14 years ago
#9
(Original post by C4>O7)
How is it shown that: 'cos(2x) ≡ cos^2(x) - sin^2(x)' ?
I think you have to look at the trigonometric proof, that should be in the P2 textbook.

(Original post by Nick Sutcliffe)
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
Surely you have just qutoed the above, where C4>O7's has A=B
0
14 years ago
#10
I'm pretty sure that is the trigonometric proof given in P2. The proof of cos(a + b) = cos(a)cos(b) - sin(a)sin(b) , I believe, is beyond pure 2.
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14 years ago
#11
(Original post by Nick Sutcliffe)
I'm pretty sure that is the trigonometric proof given in P2. The proof of cos(a + b) = cos(a)cos(b) - sin(a)sin(b) , I believe, is beyond pure 2.
u can do an algebraic proof; expand cos(a + b) and u get:

1 - [(a+b)^2]/2! + [(a+b)^4]/4! - ...

collect the terms up, and after a lot of messy algebra u will get:

(1 - [a^2]/2! + [a^4]/4! - ...)*(1 - [b^2]/2! + [b^4]/4! - ...) - (a - [a^3]/3! + [a^5]/5! - ...)*(b - [b^3]/3! + [b^5]/5! - ...)

which is also known as:

cos(a)cos(b) - sin(a)sin(b)

u can also do an analytical proof, ie with triangles and angles.
0
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