Proof of sen² 25º + cos² 45º + sen² 65º = 3/2

[Fanekaz]

Homework Statement

Show that sen² 25º + cos² 45º + sen² 65º = 3/2

Homework Equations

cos 45º= 1

The Attempt at a Solution

I get confused with the ² on the cos and sen...

 

[tiny-tim]

Hi Fanekaz!

(btw, it's sin, not sen )

I get confused with the ² on the cos and sen...

It just means that the whole thing is squared …

so sin²x means (sinx)² …

(pronounce it "sine squared of x")

it saves writing brackets. ​

 

[kerimek]

I can provide a proof for this equation using trigonometric identities. First, we can rewrite the equation as sin² 25º + cos² 45º + sin² 65º = 3/2.

Using the Pythagorean identity, we know that sin² x + cos² x = 1 for any angle x. Therefore, we can substitute sin² 25º with 1 - cos² 25º and sin² 65º with 1 - cos² 65º.

Now our equation becomes (1 - cos² 25º) + cos² 45º + (1 - cos² 65º) = 3/2.

Next, we can use the identity cos² x + sin² x = 1 to simplify the equation further. This gives us (1 - cos² 25º) + (1 - cos² 65º) = 3/2.

Expanding the brackets, we get 2 - cos² 25º - cos² 65º = 3/2.

Subtracting 2 from both sides, we are left with -cos² 25º - cos² 65º = -1/2.

Using the double angle formula for cosine, we can rewrite cos² x as (1 + cos 2x)/2. Substituting 25º and 65º for x, we get -((1 + cos 50º)/2) - ((1 + cos 130º)/2) = -1/2.

Simplifying further, we get -(2 + cos 50º + cos 130º) = -1.

Using the fact that cos 45º = 1/√2 and cos 135º = -1/√2, we can rewrite the equation as -(2 + cos 50º - cos 45º) = -1.

Simplifying again, we get -(2 + cos 50º - 1/√2) = -1.

Finally, solving for cos 50º, we get cos 50º = 1/√2, which is true. Therefore, our original equation is proven to be true.