Question

cos² π÷8+cos²3π÷8+cos²5π÷8+cos²7π÷8​

Answer 1

cos2A≡2cos²A-1 so cos²A≡½(1+cos2A).

Therefore, we can substitute for the squares of the cosines:

cos²(π/8)=½(1+cos(π/4))=½(1+√2/2);

cos²(3π/8)=½(1+cos(3π/4))=½(1-√2/2);

cos²(5π/8)=½(1+cos(5π/4))=½(1-√2/2);

cos²(7π/8)=½(1+cos(7π/4))=½(1+√2/2);

Add these together and we get the result 4/2=2 QED