find the sum cos^2(j) + sin^2(j), if j = p/3
-do I have to use a pythagorean identity?
The easiest way to proceed with this is to recognize that cos^2(x) + sin^2(x) = 1.
So, no matter what x is, even if it is p/3, the sum is 1.
Pythagorean thm works - think of the unit circle:
hypotenuse (or radius) is always 1 on the unit circle, so c^2 = a^2 + b^2 is the same as hyp^2 = cos^2 + sin^2 - right?
Thanks I was using it and getting 1 as the answer but since I have to plug in six more of the "p" fractions with different denominators I thought it didn't make since that for each one would be 1
they're just trying to drive the idea home with a hammer
your first instincts were right on target
sorry - lost my name somewhere along the way
Yes, you can use a Pythagorean identity to simplify the expression. The Pythagorean identity for trigonometric functions states that cos^2(j) + sin^2(j) = 1. Therefore, substituting j = π/3 into the expression, we have cos^2(π/3) + sin^2(π/3) = 1.
To find the value of cos(π/3) and sin(π/3), we can use the unit circle or reference angles. In this case, the angle π/3 or 60 degrees is located in the first quadrant, where cos(π/3) = 1/2 and sin(π/3) = √3/2.
Substituting these values, we have (1/2)^2 + (√3/2)^2 = 1/4 + 3/4 = 4/4 = 1. Therefore, the sum cos^2(π/3) + sin^2(π/3) equals 1.