I am having a lot of trouble proving these trig identities! If anyone can help me out, that would be fantastic!
2 csc 2x = sec x csc x
2 cot 2x = cot x - tan x
and ...
cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x ( question editted!!! )
THANKS
I am having a lot of trouble proving these trig identities! If anyone can help me out, that would be fantastic!
2 csc 2x = sec x csc x
2 cot 2x = cot x - tan x
and ...
cos^6x + sin^6x = 1 - 3 sin^2x + 3 sin^4x ( question editted!!! )
THANKS
2 csc 2x = sec x csc x
LHS = 2 csc 2x
= 2 / sin 2x
= 2 / 2 sin x cos x
= 1 / sin x cos x
= csc x sec x
= RHS
Thefore, proven.
2 cot 2x = cot x - tan x
LHS = 2 cot 2x
= 2 / tan 2x
= 2 / [2 tan x / 1 - tan^2 x]
= 2[1 -tan^2 x] / 2 tan x
= 1 - tan^2 x / tan x
= cot x -tan x
= RHS
Therefore, proven.
and ...
cos6x + sin6x = 1 - 3 sin2x + 3 sin4x
Let us check if that is true when x = 30 degrees,
cos(6*30deg) +sin(6*30deg) =? 1 -3sin(2*30deg) +3sin(4*30deg)
-1 +0 =? 1 -3(0.866) +3(0.866)
-1 =? 1
No.
So, cos6x + sin6x = 1 - 3 sin2x + 3 sin4x is not an identity.
Hey guys! I editted the last question, can you solve it now? THANKS SO MUCH!
@ Soroban: how does 1 / cos x * 1 / sin x give you 2 / 2sinxcosx
1 times 1 is 1 right? So where does the 2 come in?
@ticbol:
how do you get this step highlighted in bold?
LHS = 2 csc 2x
= 2 / sin 2x
= 2 / 2 sin x cos x
= 1 / sin x cos x
= csc x sec x
= RHS
what happens to the first 2 in 2 csc 2x?
He noted that . He only had a , so he multiplied the numerator and denominator by 2 (essentially multiplying the fraction by 1.)
You can get the same result by noting that , putting this into the denominator and then simplifying the fraction. Since many people really don't like complex fractions, I think Soroban chose the best course.
-Dan