Trig Identities Maths easy.

[1992]

Could someone do this identity with working out please
cot@ -tan@ = 2cot2@

 

[princesszahra]

open everything in sin and cos form!
and then do it u can do it !

 

[Anomaly]

cos@/sin@ - sin@/cos@
(cos^2@ - sin^2@)/sin@cos@

As the numerator is equal to cos2@ hence
cos2@/sin@cos@
Multiply numerator and denominator by 2
you get 2cos2@/2sin@cos@
(Sin2@ =2sin@cos@)
it becomes 2cos2@/sin2@ which equals RHS

 

[1992]

thanks dude
Could u do this one too please, i tried it 2 times but doesnt come out answer.

cot@ -cot2@ = cosec 2@

 

[sse2010]

change cot@ n cot 2@ in cos@/sin@ - cos2@/sin2@
=cos@/sin@-(1-2sin^2@)/2sin@cos@ (open cos 2@ n sin2@)
=(2cos^2@-1+2sin^2@)/2sin@cos@
=1/2sin@cos@ (sin^2@+cos^2@=1)
=1/sin2@
=cosec2@

 

[1992]

thanks
can somebody show this one too please:

sin^2@ cos^2@ = (1-cos4@) /8

 

[PlanetMaster]

thanks
can somebody show this one too please:

Sin^2@ Cos^2@ = (1-Cos4@) /8

= Sin²(x)Cos²(x)
= (1/4) (4Sin²(x)Cos²(x))
= (1/4) (2Sin(x)Cos(x))²
= (1/4) (Sin(2x))²
= (1/4) Sin²(2x)
= (1/4) {[1 - Cos(2.2x)]/2}
= (1/8) (1 - Cos(4x)) (Shown)