Average Boreduser
𝙱𝚊𝚗𝚗𝚎𝚍
How do you do these?
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More Trig (because I'm stupid) (1 Viewer)
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Anaya R
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Still working out the first one, but for the second one try applying the fact that cos3x = cos(x+2x) and expand that. By doing so you'll be able to apply the sin2x and cos2x double angle identities
Anaya R
Well-Known Member
Ok with the first one you want to let theta=2x. (The implication is that theta/2 = x). Doing that will give you (sin2x+sinx)/(1+cos2x+cosx). If you expand the double angle identities and collect like terms, you'll be able to cancel out some stuff to get sinx/cosx. That is equal to tanx. You can then sub in theta/2 and tan theta/2 = t.
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Average Boreduser
𝙱𝚊𝚗𝚗𝚎𝚍
Hi Thanks so much
aim for 95+
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Q2wish my handwriting is not so bad to understand.
Good luck Tomorrow
Good luck Tomorrow
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No. You're not stupid. It took me a while to do it. Not that straightforward.
You can derive, if necessary, that: 2sinAsinB = cos(A-B) - cos(A+B) from:
cos(A+B) = cosAcosB - sinAsinB and cos(A-B) = cosAcosB + sinAsinB
 - cos(4A+2A)= cos2A - cos6A \\ \\ =(2cos^2 A - 1)) - (2cos^23A - 1)\\ \\ = 2(cos^2 A - cos^2 3A) \\ \\ \therefore cos^2 \theta - cos^2 3\theta = sin 4\theta sin2\theta)
QED
You can derive, if necessary, that: 2sinAsinB = cos(A-B) - cos(A+B) from:
cos(A+B) = cosAcosB - sinAsinB and cos(A-B) = cosAcosB + sinAsinB
QED
Last edited:
5uckerberg
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For the question 
, we can start like this
+\sin{3x}+\sin\left(3x+2x\right)}{\cos\left(3x-2x\right)+\cos{3x}+\cos\left(3x+2x\right)})
.
This just becomes
Simplifying we have
Factorising it becomes
}{\cos{3x}\left(2\cos{2x}+1\right)}=\frac{\sin{3x}}{\cos{3x}}=\tan{3x})
This just becomes
Simplifying we have
Factorising it becomes