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## if (1-a^2)sin-2acos=0,prove a^2(1+cos)=1-cos

Prove a^2(1+cos)=1-cos if 1-a^2)sin - 2acos =0

Sine and Cosine are functions, so they must be performed on something.  Are these all sines and cosines of the same variable?

I agree with John absolutely!!! Konisko, you are a little too free with math functions! What you wrote is  a2(1 + c * o * s) = 1 - c * o * s
In Mathematics, there is big difference between cos(x) which is function and cos=c*o*s which is expression. You have to copy the problem precisely.

Hi Konisko!

Looks like (1-a^2)= 2a(cos/sin) or 2a(cot) ... tan = 2a / (1-a^2) = opp / adj ...

View (1-a^2) as "adj" of a right triangle with the "opp" as 2a and the h^2 is

h^2 = 4a^2 + (1 - 2a^2 + a^4) = a^4 + 2a^2 +1 ... meaning h = a^2 + 1 ...

Plug in left side: a^2= h-1 ... (h-1)(1+cos)= (h-1)(1+ adj/h) = h + adj - 1 - adj/h

Right side: 1-cos = 1 - adj/h ... leaving h + adj = 2 ... (a^2 + 1) + (1-a^2) = 2

SUMMARY: view the given "2a" and "1-a^2" as sides of a right triangle ...

Best regards  :)