Thread: 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.

Nobody still? I could use some help, to understand this atleast.

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.

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.

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.

I've done the last one.
Still didn't get you on the first one. How do i use the tan(A+B) identity?

Just write down the identity and follow your nose !

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.

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.

95 + 40 = 135 is important.

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.

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)