Trigonometric identities : Why Sin(π+t) should equal -Sin(t) ?

I'm supposed to resolve sin(π+t) and it should equal -sin(t). I'm lost

I know that :
sin(-θ) = -sinθ

and that -sinθ = cos(θ + π/2)

...and that sin(θ+α) = sinθcosα + sinαcosθ

But honestly, I'm not able to demonstrate that sin(π+t) = -sin(t)

Can anybody help me ??
thank you so much

ElectroNewby said:

I'm supposed to resolve sin(π+t) and it should equal -sin(t). I'm lost

I know that :
sin(-θ) = -sinθ

and that -sinθ = cos(θ + π/2)

...and that sin(θ+α) = sinθcosα + sinαcosθ

But honestly, I'm not able to demonstrate that sin(π+t) = -sin(t)

Can anybody help me ??
thank you so much

Click to expand...

sin(theta+pi) = sin(theta)*cos(pi)+cos(theta)*sin(pi)
sin(theta+pi) = sin(theta)*(-1) +cos(theta)*(0)
sin(theta+pi) = -sin(theta)

Ratch

Ratchit said:

sin(theta+pi) = sin(theta)*(-1) +cos(theta)*(0)

Click to expand...

Thanks a lot for the quick answer. However, I don't know how you come up with that -1 and 0 ?
The only law i know is "cos^2(theta) + sin^2(theta) = 1"

Edit: sir, I am an idiot. I overlooked what I learned previously about Pi being 180 degrees. I used the Pi function in the calculator instead of using its radian value.

However, it woke me up. Thanks a lot

Never heard of a PI function. What is it?

PI is the ratio of the circumference and the diameter of a circle. A unitless number.
There are PI Radians in 180 degrees which comes from the circumference of a circle is 2*PI*r; 360 deg in 2*PI radians hence PI radians in 180 degrees.

Calculators like to power up in radians mode and don;t like to tell you what mode it's in. Always check a known value like cosine of 180 degrees. If the cos(180) is -1, your in degree mode.

I do the following : Degree * (Pi/180) = radians
Once I have the radians, I hit the "S <-> D" key of my calculator to get the Pi radian fraction and I know that Pi is always the numerator.
I know using that key is weak but it works well