Thread: Trig Identities - Double Angles

Prove the following using double angle formulas:



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My Attempt:

Your
[sec(pi -x) / csc(2pi -x)] = 1 / [cos(pi)cos(x) +sin(pi)sin(x)]
is not correct.

[sec(pi -x) / csc(2pi -x)]
= [1 / cos(pi -x)] / [1 / sin(2pi -x)]
= sin(2pi -x) / cos(2pi -x)

So,
[sin(pi +x) / cos(2pi +x)] +[sec(pi -x) / csc(2pi -x)] = 0

LHS =
= [sin(pi +x) / cos(2pi +x)] +[sec(pi -x) / csc(2pi -x)]

= [sin(pi +x) / cos(2pi +x)] +[sin(2pi -x) / cos(pi -x)]

= [{sin(pi)cos(x) +cos(pi)sin(x)} / {cos(2pi)cos(x) -sin(2pi)sin(x)}]
+ [{sin(2pi)cos(x) -cos(2pi)sin(x)} / {cos(pi)cos(x) +sin(pi)sin(x)]

Now,
sin(pi) = 0
cos(pi) = -1
sin(2pi) = 0
cos(2pi) = 1

= [{0*cos(x) +(-1)sin(x)} / {(1)cos(x) -0*sin(x)}]
+ [{0*cos(x) -(1)sin(x)} / {(-1)cos(x) +0*sin(x)]

= [-sin(x) / cos(x)] +[-sin(x) / -cos(x)]

= -tan(x) +tan(x)

= 0

= RHS

Therefore, proven.

Hello, Macleef!

Prove: .

The first fraction is: .

. .


The second fraction is: .

. .


The problem becomes: .