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