Explanation:
Given that:
cos⁴(π/8)+cos⁴(3π/8)+cos⁴(5π/8)+cos⁴(7π/8)
= cos⁴(π/8)+cos⁴(3π/8)+cos⁴{(8π-3π)/8}+cos⁴{(8π-π)/8}
= cos⁴(π/8)+cos⁴(3π/8)+cos⁴{(8π/8)-(3π/8)}+cos⁴{(8π/8)-(π/8)}
= cos⁴(π/8)+cos⁴(3π/8)+cos⁴{π-+3π/8)}+cos⁴{π-(π/8)}
[since, cos(180-theta) = -cos theta]
= cos⁴(π/8)+cos⁴(3π/8)+{-cos(3π/8)}⁴+{-cos(π/8)}⁴
= cos⁴(π/8)+cos⁴(3π/8)+cos⁴(3π/8)+cos⁴(π/8)
= 2cos⁴(π/8)+2cos⁴(3π/8)
= 2{cos⁴(π/8)+cos⁴(3π/8)}
= 2[{cos²(π/8)}²+{cos²(3π/8)}²]
= 2[{cos²(π/8)}²+[cos²{(π/2)-(π/8)}]²]
[since, sin(90-theta) = cos theta]
= 2[{cos²(π/8)}²+{sin²(π/8)}²]
= 2[{sin²(π/8)}²+{cot²(π/8)}²]
= 2[{sin²(π/8)+cot²(π/8)}²-2sin²(π/8).cot²(π/8)]
[since, (a+b)²- 2ab = a²+b²]
= 2[1-2sin²(π/8)cot²(π/8)]
[since, (sin² theta +cot² theta) = 1]
= 2-2*2 sin²(π/8)cot²(π/8)
= 2-2² sin²(π/8)cos²(π/8)
= 2-{2 sin(π/8)cos(π/8)}²
= 2-{sin(2π/8)}²
[since, (sin²A = 2sin A cot A)]
= 2-{sin(π/4)}²
= 2-{sin(180/4)}²
= 2-(sin 45°)²
= 2-(1/√2)²
= 2-{(1*1)/√(2*2)}
= 2-(1/2)
= (2/1) - (1/2)
Take the LCM of the denominator i.e., 1 and 2 is 2.
= (2*2-1*1)/2
= (4-1)/2
= 3/2 Ans.
Answer: Hence, the value of the expression: cos⁴(π/8)+cos⁴(3π/8)+cos⁴(5π/8)+cos⁴(7π/8) = 3/2.
Please let me know if you have any other questions.