Is sin(theta) always equal to cos(pi/2-theta)?

Answer 1

Sure it does. if you think of sin and cos in terms of a ratios of the sides of a right triangle, you'll find that they correspond to the same ratios of sides. For values of theta larger than 90, it gets slightly more complicated. I think the simplest way to think about it is to realize that you get the same values for sin and cos, but with the possibility of a negative. Try 4 cases for the theta (in the 4 quadrants) and see where pi/2-theta winds up. Or you could think of pi/2 -theta as coming of the y-axis clockwise and theta as coming off the x-axis counter-clockwise (this is a lot less intuitive in my opinion.)

Answer 2

Of course it is. Draw a 3-4-5 right triangle and let A be the angle between 4 and 5 and B be the angle between 3 and 5. BY DEFINITION, sin A = 3/5, cos A = 4/5, sin B = 4/5, cos B = 3/5. So sin A and cos B are equal, and of course A+B = 90°.

If you want a proof using the difference formula, cos(π/2 - Θ) = cos(π/2)cos(Θ) + sin(π/2)sin(Θ) = 0•cos Θ + 1•sin Θ = sin Θ.

Answer 3

No, hardly ever. I think that might be a calculus identity, though, like the derivative of sin is cos and the derivative of cos is -sin, if I remember correctly.

Answer 4

as mentioned somewhere above if you are using radians then you are correct.

In radians, 1deg = 0.01745radians
so 180deg = pi radians

normally in trig, sinx = cos (90 - x)
in radians, since pi/2 = 90deg, then sinx = cos (pi/2 - x)

so your statement is true but only in radians. have fun with math!

Answer 5

If you are measuring in radians, yes. To check, use the cosine angle difference formula.

Answer 6

NO! Are you crazy?????