Trig proof Q (1 Viewer)

[5647382910]

im using @ for theta

prove the following:

sin(n@ + @) + 2sin(n@) + sin(n@ - @) = cot(@/2)
cos(n@ - @) - cos(n@ + @)

I can simplify the expression down so that:
LHS =
cos@ + 1
sin@

but i cant go any further

thanks in advance

 

[lyounamu]

im using @ for theta

prove the following:

sin(n@ + @) + 2sin(n@) + sin(n@ - @) = cot(@/2)
cos(n@ - @) - cos(n@ + @)

I can simplify the expression down so that:
LHS =
cos@ + 1
sin@

but i cant go any further

thanks in advance

use t formula
say: let t = tan(@/2)

so (cos@+1)/sin@ = ((1-t^2)+(1+t^2))/2t I skipped many steps here
= 2/2t = 1/t = 1/tan(@/2) = cot(@/2)

 

[Trebla]

The fact that the LHS has @ and the RHS has @/2, indicates that you should use the half angle results or t-formulae.

 

[5647382910]

LOL... that freakin q was in the products as sums and differences excercise, the half angle exercise was next
aha, thanks 4 the help

 

[lyounamu]

LOL... that freakin q was in the products as sums and differences excercise, the half angle exercise was next
aha, thanks 4 the help

yeah, you can go further without t-formula though.

(cos@+1)/sin@ = ((2cos^2(@/2) - 1)+1)/(2sin(@/2)cos(@/2))
= (2cos^2(@/2))/(2sin(@/2)cos(@/2))
= cos(@/2)/sin(@/2) = cot(@/2)

 

[5647382910]

yeah, you can go further without t-formula though.

(cos@+1)/sin@ = ((2cos^2(@/2) - 1)+1)/(2sin(@/2)cos(@/2))
= (2cos^2(@/2))/(2sin(@/2)cos(@/2))
= cos(@/2)/sin(@/2) = cot(@/2)

ahhk... didnt see that, thanks for the help again