Let's work on the left side first. And remember that
the tangent is the same as sin/cos.

sin(a) cos(a) tan(a)

Substitute for the tangent:

[ sin(a) cos(a) ] [ sin(a)/cos(a) ]

Cancel the cos(a) from the top and bottom, and you're left with

[ sin(a) ] . . . . . [ sin(a) ] which is [ sin²(a) ] That's the left side.

Now, work on the right side:

[ 1 - cos(a) ] [ 1 + cos(a) ]

Multiply that all out, using FOIL:

[ 1 + cos(a) - cos(a) - cos²(a) ]

= [ 1 - cos²(a) ] That's the right side.

Do you remember that for any angle, sin²(b) + cos²(b) = 1 ?
Subtract cos²(b) from each side, and you have sin²(b) = 1 - cos²(b) for any angle.

So, on the right side, you could write [ sin²(a) ] .

Now look back about 9 lines, and compare that to the result we got for the left side .

They look quite similar. In fact, they're identical. And so the identity is proven.

Whew !

4 0

3 years ago

Read 2 more answers

a quadrilateral has angles measuring 105 degrees, 80 degrees, and 45 degrees what is the measure of the 4th angle

Bond [772]

Answer: 130 degrees

Step-by-step explanation:

105+80+45+x=360

230+x=360

-230 both sides

X=130 degrees

6 0