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Reply 20
Original post by SophieL1996
2:smile:


So 2sin2θ+2cos2θ=22sin^2\theta + 2cos^2\theta =2 now you can simplify it fully
Reply 21
Original post by Robbie242
So 2sin2θ+2cos2θ=22sin^2\theta + 2cos^2\theta =2 now you can simplify it fully


2(sin^2\theta + cos^2\theta)=2 ? apparently the answer is 4 :smile:
Reply 22
Original post by SophieL1996
2(sin^2\theta + cos^2\theta)=2 ? apparently the answer is 4 :smile:
Yes, remember the 2 from earlier? add that on :smile:
Original post by SophieL1996
2(sin^2\theta + cos^2\theta)=2 ? apparently the answer is 4 :smile:


Use to help in your posts ... look at mine and Robbie's

Go back to your original question and see where this 2 fits in
Reply 24
Original post by Robbie242
Yes, remember the 2 from earlier? add that on :smile:


ah yes so it does =4, how come you add rather than subtract, as it was on the LHS and we are moving it to the right ? :biggrin:
Reply 25
Original post by SophieL1996
ah yes so it does =4, how come you add rather than subtract, as it was on the LHS and we are moving it to the right ? :biggrin:


In this case no, we are just simplifying 2+2sin2θ+2cos2θ2+2sin^2\theta+2cos^2\theta which you can substitute 2 in instead of the 2sin2θ+2cos2θ2sin^2\theta + 2cos^2\theta
(edited 11 years ago)
Reply 26
Original post by Robbie242
In this case no, we are just simplifying 2+2sin2θ+2cos2θ2+2sin^2\theta+2cos^2\theta which you can substitute 2 in instead of the [ex]2sin^2\theta + 2cos^2\theta


Ok thanks! :biggrin:
Reply 27
Original post by Robbie242
In this case no, we are just simplifying 2+2sin2θ+2cos2θ2+2sin^2\theta+2cos^2\theta which you can substitute 2 in instead of the 2sin2θ+2cos2θ2sin^2\theta + 2cos^2\theta


could you help me with this one please? :smile: sin4θ+sin2θcos2θsin^4\theta+sin^2\theta cos^2\theta
Reply 28
Original post by SophieL1996
could you help me with this one please? :smile: sin4θ+sin2θcos2θsin^4\theta+sin^2\theta cos^2\theta


Using the rule sin2θ+cos2θ=1sin^2\theta+cos^2\theta=1 appreciate that cos2θcos^2\theta can be expressed as 1sin2θ1-sin^2\theta
Now you can get terms purely in terms of sin and that will help you cancel out terms as seen later

note that later you'll see sin4θ=(sin2θ)(sin2θ)-sin^4\theta=(-sin^2\theta)(sin^2\theta)
Reply 29
Original post by Robbie242
Using the rule sin2θ+cos2θ=1sin^2\theta+cos^2\theta=1 appreciate that cos2θcos^2\theta can be expressed as 1sin2θ1-sin^2\theta
Now you can get terms purely in terms of sin and that will help you cancel out terms as seen later

note that later you'll see sin4θ=(sin2θ)(sin2θ)-sin^4\theta=(-sin^2\theta)(sin^2\theta)


I got to the point sin^4 theta + sin^2 theta - 2sin^2theta then I put it as sin^4 theta - sin^2 theta. where do I go from there?
Reply 30
Original post by SophieL1996
I got to the point sin^4 theta + sin^2 theta - 2sin^2theta then I put it as sin^4 theta - sin^2 theta. where do I go from there?



sin4θ+sin2θcos2θ=sin4θ+sin2θ(1sin2θ)sin^4\theta+sin^2\theta cos^2\theta=sin^4\theta+ sin^2\theta (1-sin^2\theta) Expand and simplify. Hint: the sin4θsin^4\theta will cancel out.

It's not 2sin2θ2sin^2\theta purely because there's not a coefficient greater than 1 involved in the expansion of the bracket
(edited 11 years ago)
Reply 31
Original post by Robbie242
sin4θ+sin2θcos2θ=sin4θ+sin2θ(1sin2θ)sin^4\theta+sin^2\theta cos^2\theta=sin^4\theta+ sin^2\theta (1-sin^2\theta) Expand and simplify. Hint: the sin4θsin^4\theta will cancel out.

It's not 2sin2θ2sin^2\theta purely because there's not a coefficient greater than 1 involved in the expansion of the bracket


Oh yes thank you!!
Reply 32
Original post by SophieL1996
Oh yes thank you!!
No problem :biggrin: anytime

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