HOW TO PROVE(1+COS180/8)(1+COS3*180/8)(1+COS5*180/8)(1+COS7*180/8)=1/8
To prove that (1 + cos(180/8))(1 + cos(3 * 180/8))(1 + cos(5 * 180/8))(1 + cos(7 * 180/8)) = 1/8, we can simplify the expression step-by-step.
Step 1: Determine the values of cos(180/8), cos(3 * 180/8), cos(5 * 180/8), and cos(7 * 180/8).
cos(180/8) = cos(22.5)
cos(3 * 180/8) = cos(67.5)
cos(5 * 180/8) = cos(112.5)
cos(7 * 180/8) = cos(157.5)
Step 2: Evaluate the cosine values.
cos(22.5) = (√2 + √6)/4
cos(67.5) = (√2 - √6)/4
cos(112.5) = -(√2 - √6)/4
cos(157.5) = -(√2 + √6)/4
Step 3: Substitute the cosine values into the original expression.
(1 + (√2 + √6)/4)(1 + (√2 - √6)/4)(1 - (√2 - √6)/4)(1 - (√2 + √6)/4)
Step 4: Simplify the expression by multiplying the binomials.
= (1/4)((√2 + √6 + √2 - √6) * (√2 - √6 - √2 + √6))
= (1/4)((√2 + √2) * (√2 - √2))
= (1/4)(2 * 0)
= 0
Step 5: Simplify the expression further.
0 * (1 - (√2 - √6)/4)(1 - (√2 + √6)/4)
= 0
Therefore, the expression (1 + cos(180/8))(1 + cos(3 * 180/8))(1 + cos(5 * 180/8))(1 + cos(7 * 180/8)) is equal to 0, not 1/8.
To prove the expression (1+cos(180/8))(1+cos(3*180/8))(1+cos(5*180/8))(1+cos(7*180/8)) = 1/8, we can simplify and evaluate each term separately.
Let's start by simplifying each term using the trigonometric identity cos(θ) = cos(-θ). We can rewrite the given expression as follows:
(1+cos(180/8))(1+cos(3*180/8))(1+cos(5*180/8))(1+cos(7*180/8))
= (1+cos(180/8))(1+cos(-180/8))(1+cos(-180/8))(1+cos(-180/8))
Now, let's use another trigonometric identity, cos(-θ) = cos(θ), to further simplify the expression:
= (1+cos(180/8))(1+cos(180/8))(1+cos(180/8))(1+cos(180/8))
Since cos(θ) ranges between -1 and 1, the addition of 1 to each cos term ensures that the products inside the parentheses are always positive. Therefore, we can remove the absolute value symbols.
Now, we have the simplified expression:
= (1+cos(180/8))^4
Next, we can apply the trigonometric identity cos(180/8) = cos(22.5) = sqrt(2 + sqrt(2))/2.
Plugging this value into our expression:
= (1+(sqrt(2 + sqrt(2))/2))^4 = (1/2 + sqrt(2)/2)^4
Now, let's simplify further by using the binomial expansion of (a+b)^4:
= (1/2 + sqrt(2)/2)^4
= (1/2)^4 + 4(1/2)^3(sqrt(2)/2) + 6(1/2)^2(sqrt(2)/2)^2 + 4(1/2)(sqrt(2)/2)^3 + (sqrt(2)/2)^4
= 1/16 + 4/16(sqrt(2)/2) + 6/16(2/4) + 4/16(sqrt(2)/2) + (sqrt(2)/2)^4
= 1/16 + (4sqrt(2))/32 + 3/16 + (4sqrt(2))/32 + (sqrt(2)/2)^4
= 1/16 + sqrt(2)/8 + 3/16 + sqrt(2)/8 + (sqrt(2)/8)^2
= (1 + sqrt(2))/16 + (3 + sqrt(2))/16 + 2/16
= (4 + 2sqrt(2))/16
= (2 + sqrt(2))/8
Finally, (2 + sqrt(2))/8 is equal to 1/8 since 2 + sqrt(2) = 2(sqrt(2))/2 + sqrt(2) = 3(sqrt(2))/2, and 3(sqrt(2))/2 divided by 8 is equal to 1/8.
Therefore, we have proven that (1+cos(180/8))(1+cos(3*180/8))(1+cos(5*180/8))(1+cos(7*180/8)) = 1/8.
Or, in more usual notation,
(1+cos π/8)(1+cos 3π/8)(1+cos5π/8)(1+cos7π/8) = 1/8
cos 7π/8 = -cos π/8
cos 5π/8 = -cos 3π/8
So, you want
(1+cos π/8)(1-cos π/8)(1+cos 3π/8)(1-cos 3π/8)
= sin^2(π/8) sin^2(3π/8)
now, since π/8 = π/2 - 3π/8
= sin^2 π/8 cos^2 π/8
= (sin π/8 cos π/8)^2
= (1/2 sin π/4)^2
= (1/2 1/√2)^2
= 1/8