prove that sin sqared pie/6+ cos sqared pie/3-tansqare pie/4=1/2 in trigonometry
GRR! PI, not pie!
sin pi/6 = cos pi/3 = 1/2
tan pi/4 = 1
so, we have
1/4 + 1/4 - 1 = -1/2
I suspect a typo
To prove that sin^2(pi/6) + cos^2(pi/3) - tan^2(pi/4) = 1/2 in trigonometry, we need to simplify each trigonometric term and show that they add up to 1/2.
Let's break down each term and simplify them one by one:
1. sin^2(pi/6):
Using the Pythagorean identity, sin^2(theta) + cos^2(theta) = 1, we can express sin^2(pi/6) in terms of cos(pi/6):
sin^2(pi/6) = 1 - cos^2(pi/6)
2. cos^2(pi/3):
Similarly, using the Pythagorean identity, sin^2(theta) + cos^2(theta) = 1, we can express cos^2(pi/3) in terms of sin(pi/3):
cos^2(pi/3) = 1 - sin^2(pi/3)
3. tan^2(pi/4):
Recall that tan(theta) = sin(theta) / cos(theta). Substituting the values of sin(pi/4) and cos(pi/4), we can express tan^2(pi/4) as:
tan^2(pi/4) = [sin(pi/4) / cos(pi/4)]^2
Now, let's substitute these values into the equation and simplify:
sin^2(pi/6) + cos^2(pi/3) - tan^2(pi/4)
= (1 - cos^2(pi/6)) + (1 - sin^2(pi/3)) - [sin(pi/4) / cos(pi/4)]^2
Next, we can simplify the equation using the values of sin(pi/6), cos(pi/6), sin(pi/3), cos(pi/3), sin(pi/4), and cos(pi/4):
= (1 - (1/2)^2) + (1 - (sqrt(3)/2)^2) - [(sqrt(2)/2) / (sqrt(2)/2)]^2
= (1 - 1/4) + (1 - 3/4) - (1^2)
= (3/4) + (1/4) - 1
= 1/2 - 1
= -1/2
As we can see, the simplified equation is not equal to 1/2, which means the initial equation is not true.
Hence, we have disproven the equality sin^2(pi/6) + cos^2(pi/3) - tan^2(pi/4) = 1/2 in trigonometry.