# Help needed

• Oct 4th 2012, 06:56 AM
omardalvi
Help needed
These are the questions
1) Evaluate : tan 95 + tan 40 - tan 95 * tan 40

2) If (√8 + i)^50 = 3^49(a+ib), find a^2 and b^2

3) Prove that : 2cos(pi/13)cos(9pi/13)+cos(3pi/13)+cos(5pi/13)=0

One is from complex numbers. I'm in XIth, need help, i have the formulas, but i can't solve it ,unfortunately. Someone explain it, or atleast solve it please.
And if anyone can give related questions for me to practise, i don't mind.
Thanks in advance. I've been trying it the whole day, and my its doing my head in.
• Oct 4th 2012, 08:38 AM
omardalvi
Re: Help needed
Nobody still? I could use some help, to understand this atleast.
• Oct 4th 2012, 09:02 AM
HallsofIvy
Re: Help needed
Perhaps we do not understand what you mean by "help". You have posted several problems but not shown what you have tried yourself or what you know about the problems.
• Oct 4th 2012, 09:08 AM
omardalvi
Re: Help needed
By help i mean i need help to solve these problems. I've tried applying the identities but i couldn't get the final answer. I need to know how to solve them.
Especially the first one.
• Oct 5th 2012, 04:28 AM
BobP
Re: Help needed
For the first one, use the trig identity for tan(A+B) and notice that 95+4.
For the second one, the general advice is that to raise a complex number to some power, first switch to polars.
For the third one make use of the identity for cosA + cosB.
• Oct 5th 2012, 04:43 AM
omardalvi
Re: Help needed
I've done the last one.
Still didn't get you on the first one. How do i use the tan(A+B) identity?
• Oct 5th 2012, 05:02 AM
BobP
Re: Help needed
• Oct 5th 2012, 05:10 AM
BobP
Re: Help needed
BTW I've just noticed that in an earlier post a 0 = 135 got chopped from the end of a line.
The line should have finished as ........ 95 + 40 = 135.
• Oct 5th 2012, 06:09 AM
omardalvi
Re: Help needed
but the identity is tan(a+b) = (tan a + tan b)/1-tan a *tan b
There is no 1-tan a tan b in denominator, and i tried adding it, and it didn't work.
• Oct 6th 2012, 02:39 AM
BobP
Re: Help needed
95 + 40 = 135 is important.
• Oct 7th 2012, 05:53 AM
StefanTM
Re: Help needed
Hi,
use 95+40=135 ==> 95=135-40
tan(135-40) + tan (40) - tan(135-40)*tan(40) = ...

tan135 = -1, in the expression will be only tan(40), not tan (95) more.(Wink)
• Oct 7th 2012, 06:18 AM
StefanTM
Re: Help needed
2) If (√8 + i)^50 = 3^49(a+ib), find a^2 and b^2
For the begin 2): 3^50((√8/3+1/3*i)^50 = 3^49(a+ib)
√8/3 = cos(fi), 1/3 =sin (fi)

3) Prove that : 2cos(pi/13)cos(9pi/13)+cos(3pi/13)+cos(5pi/13)=0
Use identity: 2cos(x)*cos(y) = cox(x+y)+cos(x-y)