Question

prove the identity: cosx (cot x sin x- tan x sin x)=1 - 2 sin^2x. Should i change it all to cos and sin? or distribute cosx?

Answer 1

change the cotx and the tanx first. it'll cancel things out and might simplify it for ya

Answer 2

If you replace cos with A/H, cot with A/O, sin with O/H, etc., it reduces to simple algebra. Canceling produces:
cos^2 - sin^2 = 1 - 2 sin^2
cos^2 = 1 - sin^2

Answer 3

cos(x)((cos(x)(sin(x))/sin(x)) - sin(x)(sin(x)/cos(x))
= cos^2(x) - sin^2(x)

Answer 4

use the rule sin(x) + cos(x) = 1 to elimante the cos^2(x) and you will find that you have the solution to your problem. I actually miss these problems they were soo easy

Answer 5

sorry

sin^(2)(x) + cos^(2)(x) = 1

Answer 6

thus cos^(2)(x) = 1 - sin^(2)(x)
so we just replace the cos with 1 - sin^(2)(x) and we end up with
1- sin^(2)(x) - sin^2(x)
= 1-2sin^(2)(x)

Answer 7

i dont understand how you are suppose to add everything inside with out common denominators?

Answer 8

inside the parenthese i mean

Answer 9

Don't add. The original equation is, by definitions of the trig terms,
(A/H) ( (A/O) (O/H) - (O/A) (O/H) ) = 1 - 2 (O/H) (O/H)
The left-hand side reduces by distributing the (A/H) and canceling to
A^2/H^2 - O^2/H^2
This is cos^2 - sin^2.
Then see my solution above.

Answer 10

you don't add anything inside, you cancel out the fractions by converting
cot(x) = cos(x)/sin(x)
tan(x) = sin(x)/cos(x)

so you have
sin(x)cos(x)/sin(x) - sin(x)sin(x)/cos(x)
sin(x)cos(x)/sin(x)
sin(x)/sin(x) = 1 so you are left with cos(x) - sin(x)sin(x)/cos(x)

Answer 11

Now distribute the outside term
cos(x)(cos(x) - sin(x)sin(x)/cos(x))
=
cos(x)cos(x) - sin(x)sin(x)cos(x)/cos(x)
cos(x)/cos(x) = 1

thus

cos(x)cos(x) - sin(x)sin(x)

Answer 12

I recommend learning the basic laws of division before tackling these problems

Answer 13

not to patronize

Answer 14

After you turn Cotx and Tanx and distribute the cosx we get:

\[cosx((cosx/sinx)sinx - (sinx/cosx)sinx) = cosx(cosx - (\sin^2x)/cosx)\]
\[= \cos^2x - (Sin^2x)\]

remember that \[\cos^2x = 1 - sin^2x\]
so we can just plug that in for our \[\cos^2x\]

and we get:\[1- \sin^2x - \sin^2x \]

combining the two sin^2x

wet get
\[ 1 - 2\sin^2x\]

ta da! now go study :)

Answer 15

haha oh man i suck at these!

Answer 16

what are you having problems with specifically? Do you agree that