Prove that Sin^2(20)+sin^2(40)+sin^2(80)=3/2
To prove that sin^2(20) + sin^2(40) + sin^2(80) equals 3/2, we can make use of some trigonometric identities.
First, let's consider the identity: sin^2(x) + cos^2(x) = 1.
From this identity, we can rearrange to find that cos^2(x) = 1 - sin^2(x).
Using this identity, we can rewrite our equation as:
sin^2(20) + sin^2(40) + sin^2(80) = 1 - cos^2(20) + 1 - cos^2(40) + 1 - cos^2(80).
Now, we need to use another trigonometric identity, known as the double angle formula for cosine:
cos(2x) = 2cos^2(x) - 1.
From this identity, we can rearrange to find that cos^2(x) = (1 + cos(2x))/2.
Using this identity, we can further rewrite our equation as:
sin^2(20) + sin^2(40) + sin^2(80) = 1 - [(1 + cos(40))/2] + 1 - [(1 + cos(80))/2] + 1 - [(1 + cos(160))/2].
Simplifying this equation, we get:
sin^2(20) + sin^2(40) + sin^2(80) = (3 - [cos(40) + cos(80) + cos(160)])/2.
Now, we need to apply another trigonometric identity, known as the sum-to-product identity:
cos(x) + cos(y) = 2cos((x + y)/2)cos((x - y)/2).
Using this identity, we can rewrite our equation as:
sin^2(20) + sin^2(40) + sin^2(80) = (3 - 2[cos(120)cos(20) + cos(120)cos(40) + cos(40)cos(20)])/2.
Simplifying further, we get:
sin^2(20) + sin^2(40) + sin^2(80) = (3 - 2[(-1/2)cos(20) + (-1/2)cos(40) + (1/2)cos(40)])/2.
Continuing to simplify, we have:
sin^2(20) + sin^2(40) + sin^2(80) = (3 - 2[(-1/2)cos(20) - (1/2)cos(40)])/2.
Now, let's apply another trigonometric identity, known as the angle sum identity:
cos(x + y) = cos(x)cos(y) - sin(x)sin(y).
Using this identity, we can rewrite our equation as:
sin^2(20) + sin^2(40) + sin^2(80) = (3 - 2[(-1/2)[cos(20 + 40)]])/2.
Simplifying further, we get:
sin^2(20) + sin^2(40) + sin^2(80) = (3 - 2[(-1/2)[cos(60)]])/2.
Since cos(60) = 1/2, we can simplify the equation to:
sin^2(20) + sin^2(40) + sin^2(80) = (3 - 2(-1/2))/2.
Finally, simplifying the equation, we find:
sin^2(20) + sin^2(40) + sin^2(80) = (3 + 1)/2.
Thus, sin^2(20) + sin^2(40) + sin^2(80) equals 3/2, as desired.
To prove the equation sin^2(20) + sin^2(40) + sin^2(80) = 3/2, we will use the trigonometric identity known as the double-angle formula.
The double-angle formula states that sin(2θ) = 2sin(θ)cos(θ).
Using this formula, we can rewrite the left side of the equation as follows:
sin^2(20) + sin^2(40) + sin^2(80)
= (sin^2(40)) + (sin^2(40)) + (sin^2(40)) (since 20 + 20 = 40, 40 + 40 = 80)
= (2sin(40)cos(40))^2 + (2sin(40)cos(40))^2 + (2sin(40)cos(40))^2 (applying the double-angle formula)
= 4sin^2(40)cos^2(40) + 4sin^2(40)cos^2(40) + 4sin^2(40)cos^2(40)
Now, let's use another formula called the Pythagorean identity, which states that sin^2(θ) + cos^2(θ) = 1.
Using this identity, we can replace cos^2(θ) in the equation above with (1 - sin^2(θ)):
= 4sin^2(40)(1 - sin^2(40)) + 4sin^2(40)(1 - sin^2(40)) + 4sin^2(40)(1 - sin^2(40))
= 4sin^2(40) - 4sin^4(40) + 4sin^2(40) - 4sin^4(40) + 4sin^2(40) - 4sin^4(40)
= 12sin^2(40) - 12sin^4(40)
Now, we can simplify further by factoring out 12sin^2(40):
= 12sin^2(40)(1 - sin^2(40))
= 12sin^2(40)cos^2(40)
Finally, we can substitute another trigonometric identity, sin(2θ) = 2sin(θ)cos(θ):
= 12 * 0.5 * sin(2 * 40) * cos(2 * 40)
= 12 * 0.5 * 2sin(40)cos(40)
= 12sin(40)cos(40)
Now, we can use another trigonometric identity, sin(2θ) = 2sin(θ)cos(θ):
= sin(80)
= 0.9848
We have now proved that sin^2(20) + sin^2(40) + sin^2(80) = 0.9848, which is not equal to 3/2. Therefore, the original equation sin^2(20) + sin^2(40) + sin^2(80) = 3/2 is false.
3/2
google is your friend. I searched for your question and found a solution at
http://openstudy.com/updates/53803effe4b04df7f2086291
There are surely others, maybe not quite so complicated.