Simplifying a trig expression

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The question wants me to simplify (sin^2 theta - cos^2 theta) - sin^2 2theta
I can siplify it a bit, but not the whole way. Can someone point me in the right direction?
 
You say you've simplified "a bit". How far have you gotten?

You say you can't simplify "all the way". Does this mean that you are heading for some intended (that is, known) expression? If so, what is it?

I'll be glad to "point [you] in the right direction", but I'll need to know where you are. Please reply with a complete listing of what you've done so far. Thank you.

Eliz.

P.S. For convenience, you can use "@" for "theta". It's quicker, and I think we all know what is meant by this. Thank you! :D
 
I just realized something... It says (sin^2@-cos^2@)^2 -sin^2 2@. I hadn't seen that extra ^2 before.
It's made it easier, but I'm still not sure how to get the right answer.
(sin^2@-cos^2@)^2 -sin^2 2@ =
sin^4@ -sin^2@cos^2@ -sin^2@cos^2@ +cos^4@ -sin^2 2@ =
sin^4@ +cos^4@ -2sin^2@cos^2@ -sin^2 2@ =
-cos4@ -sin^2 2@ -sin^2 2@ =
-cos4@ -2sin^2 2@ =
?
Where do I go from here? Or have I messed up somewhere? The answer I'm supposed to get is cos4@, but I'm not sure I can get there from here.
 
G'day BW52,

If it helps, you can write (sin^2(@)-cos^2(@))^2 as (-(cos^2(@)-sin^2(@))^2, and then use the identity cos(2@) = cos^2(@)-sin^2(@).

This makes the squaring nice. You can then use the identity again to arrive at the desired answer.
 
But then I get (-cos^2 2@)^2 -sin^2 2@ not (-cos^2 2@) -sin^2 2@ so I can't use that identity, can I?

PS. If nobody manages to get me seeing this straight tonight, I'll ask about this in class.
 
cos(2@) = cos^2(@) - sin^2(@)

not
cos^2(2@) = cos^2(@) - sin^2(@)

And you aren't looking for
(-cos^2 2@) -sin^2 2@

You're looking to get
cos^2 (2@) -sin^2 (2@)

(The square of a negative number is positive.)
 
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