(a) Try using the product-to-sum formula
cos(a)cos(b) = 1/2 (cos(a-b)+cos(a+b))
(b) We have
tan x/2 = (1-cosx)/sinx
tan 15 = (1-cos30)/sin30 = (1-√3/2)/(1/2) = 2-√3
sin15 = √((1-cos30)/2) = √((1-√3/2)/2) = √(2-√3)/2
sin15 = √((1+cos30)/2) = √((1+√3/2)/2) = √(2+√3)/2
Now apply that again, and note that
(√2+√6)^2 = 4(2+√3)
and I think things will fall out as you desire.
So I have two questions that have been puzzling me for quite some time and would really appreciate any help with either of them!
(a) There are four positive intergers a, b, c, and d such that 4cos(x)cos(2x)cos(4x)=cos(ax)+cos(bx)+cos(cx)+cos(dx) for all values of x. Find a+b+c+d.
I started this problem by trying to use the double-angle formulas to expand cos(2x) and cos(4x). This quickly seemed to become difficult as I was working with exponents and there didn't seem to be an easy way to simplify
(b) There are intergers a, b, c, and d such that tan(7.5)=sqrt(a)+sqrt(b)-sqrt(c)-sqrt(d). Find a+b+c+d.
I started this problem by using the fact that tan(x)=sin(x)/cos(x). Next I used the half-angle identities and sum-difference formulas to get a very ugly fraction that I had no idea how to solve.
Any help would be much appreciated - thank you in advance!
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Thank you so much for your help! I got both of those now!
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