Trig identites Question (1 Viewer)

[mtsmahia]

Hi can someone explain how to do this question


http://community.boredofstudies.org/attachment.php?attachmentid=18547&stc=1&d=1241065952

thanks for help!

 

[gcchick]

Change the (cos^2)b to (1 - (sin^2)b) and (cos^2)a to (1-(sin^2)a)

 

[mtsmahia]

Change the (cos^2)b to (1 - (sin^2)b) and (cos^2)a to (1-(sin^2)a)

what do u do after that?

 

[Drongoski]

 

[gcchick]

Expand the brackets.

Not posting a solution.

 

[mtsmahia]

ok...i get ti now...thanks for the help guys

 

[gcchick]

No worries.

 

[mtsmahia]

sorry....can u please answer one more question
CodeCogsEqn(4).gif

 

[Timothy.Siu]

LHS=sin³x/cos x + cos³x/sin x + 2sinxcosx

=[sin⁴x+cos⁴x+2cos²xsin²x ] / cosxsinx

= (sin²+cos²)² / sinx cosx

=1/sinxcosx
= (sin²x+cos²x)/sinx.cosx
=tanx+cotx=RHS