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Urgent help please Trigonometry a Level

Please help me... I'm stuck
Thank you :smile:
cos^4 theta + 2sin^2 theta + sin^4 theta
So far I've done:
Cos^2 theta (cos^2 theta +1) + sin^2 theta ( 2+ sin^2)
(Cos ^2 theta)( sin^2 theta) = (2+sin^2)(cos^2 theta +1)
Cos^2 sin^2 theta= 2cos^2 theta + 2 + sin^2 cos ^2 theta + sin^2
Cos^2sin^2 theta= 2cos^4 2theta + sin^4 +2
Cos^2 2 theta + sin^2 theta +2
Cos^2 2 theta + sin^2 theta + 2 = 1
Answer is 1
(edited 6 years ago)
Can you explain your first line? There's issues beyond that but your initial assertion (?) of cos⁡4θ+2sin⁡2θ+sin⁡4θ≡cos⁡2θ(cos⁡2θ+1)+sin⁡2θ(2+sin⁡2θ)\cos^4 \theta + 2\sin^2\theta +\sin^4\theta \equiv \cos^2\theta(\cos^2\theta+1) + \sin^2\theta(2+\sin^2\theta) is incorrect. Perhaps you added a +1 in the first brackets by mistake? I'm not sure where you're going with the rest.

Spoiler

(edited 6 years ago)
Can't move on the app, will move when I get home :smile:
Reply 3
Avatar for Musicanor
Musicanor
OP
9
Original post by _gcx
Can you explain your first line? There's issues beyond that but your initial assertion (?) of cos⁡4θ+2sin⁡2θ+sin⁡4θ≡cos⁡2θ(cos⁡2θ+1)+sin⁡2θ(2+sin⁡2θ)\cos^4 \theta + 2\sin^2\theta +\sin^4\theta \equiv \cos^2\theta(\cos^2\theta+1) + \sin^2\theta(2+\sin^2\theta) is incorrect. Perhaps you added a +1 in the first brackets by mistake? I'm not sure where you're going with the rest.

Spoiler



Oops I think I wrote it wrong
The question is:
Simplify the following expressions:
Cos^4 theta + 2sin^2 theta cos^2 theta + sin^4 theta
Original post by Musicanor
Oops I think I wrote it wrong
The question is:
Simplify the following expressions:
Cos^4 theta + 2sin^2 theta cos^2 theta + sin^4 theta


In that case your final answer is correct, but you have some mistakes in your working. (I'm not sure how you reached your conclusion from your last few lines. If you graph y=cos⁡22x+sin⁡2x+2y=\cos^2 2x + \sin^2x + 2 you'll see that it's neither constant nor equal to 1) It's easier to factorise. We have (cos⁡2θ)2+2sin⁡2θcos⁡2θ+(sin⁡2θ)2(\cos^2\theta)^2 + 2\sin^2\theta\cos^2\theta + (\sin^2\theta)^2. Does this look familiar?
Reply 5
Avatar for Musicanor
Musicanor
OP
9
Original post by _gcx
In that case your final answer is correct, but you have some mistakes in your working. (I'm not sure how you reached your conclusion from your last few lines. If you graph y=cos⁡22x+sin⁡2x+2y=\cos^2 2x + \sin^2x + 2 you'll see that it's neither constant nor equal to 1) It's easier to factorise. We have (cos⁡2θ)2+2sin⁡2θcos⁡2θ+(sin⁡2θ)2(\cos^2\theta)^2 + 2\sin^2\theta\cos^2\theta + (\sin^2\theta)^2. Does this look familiar?

mmmm...
cos 2 θ − sin 2 θ = 1 − 2 sin 2 θ ????
Original post by Musicanor
mmmm...
cos 2 θ − sin 2 θ = 1 − 2 sin 2 θ ????


No, that's not true, (that is, it isn't an identity) unless you mean cos⁡2\cos^2 etc.

Try expanding (x2+y2)2(x^2+y^2)^2. What do you see?
(edited 6 years ago)
Reply 7
Avatar for Musicanor
Musicanor
OP
9
Original post by _gcx
No, that's not true, (that is, it isn't an identity) unless you mean cos⁡2\cos^2 etc.

Try expanding (x2+y2)2(x^2+y^2)^2. What do you see?

X^4 +y^4 + x^4y^4
Reply 8
Avatar for Musicanor
Musicanor
OP
9
Original post by Musicanor
X^4 +y^4 + x^4y^4

Or is it 2x^4 +2x^4 2y^4 +2y^4
Original post by Musicanor
X^4 +y^4 + x^4y^4


You should be comfortable with this,

(x2+y2)2≡(x2)2+2x2y2+(y2)2≡x4+2x2y2+y4(x^2 + y^2)^2 \equiv (x^2)^2 + 2x^2y^2 + (y^2)^2 \equiv x^4 + 2x^2y^2 + y^4

You have sin⁡4θ+2sin⁡2θcos⁡2θ+cos⁡4θ\sin^4\theta + 2\sin^2\theta\cos^2\theta + \cos^4\theta
Reply 10
Avatar for Musicanor
Musicanor
OP
9
Original post by _gcx
You should be comfortable with this,

(x2+y2)2≡(x2)2+2x2y2+(y2)2≡x4+2x2y2+y4(x^2 + y^2)^2 \equiv (x^2)^2 + 2x^2y^2 + (y^2)^2 \equiv x^4 + 2x^2y^2 + y^4

You have sin⁡4θ+2sin⁡2θcos⁡2θ+cos⁡4θ\sin^4\theta + 2\sin^2\theta\cos^2\theta + \cos^4\theta

Thanks.

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