I did this a little differently (the answer is the same although Chris's is not quite finished.)
The co in cosine stands for complement of sine. Complementary angles add to 90 degrees.
So
$$\boxed{sin\theta=cos(90-\theta)}\\\\$$
also, 140 degrees is in the 2nd quadrant. Sine is positive in the 2nd quad. so
$$\\sin140^0=sin(180-140)=sin40\\\\
sin40=cos50\\
so \\
sin140=cos50$$
This is the same as CPhill's answer.
Now we know that sin50=p=p/1
Draw a right angled triangle and label on of the acute angles as 50 degrees.
The opposite side is p
The hypotenuse is 1
So using pythagoras' Theorum the adjacent side must be $$\sqrt{1-p^2}$$
$$\\cos50=\frac{adj}{hyp}=\frac{\sqrt{1-p^2}}{1}\\\\
cos50^0=\sqrt{1-p^2}$$
Using an additive identity (and assuming degrees), we have
sin 140 =
sin(90 + 50) =
sin90cos50 + sin50cos90 =
1*cos50 + p*0 =
cos50
I did this a little differently (the answer is the same although Chris's is not quite finished.)
The co in cosine stands for complement of sine. Complementary angles add to 90 degrees.
So
$$\boxed{sin\theta=cos(90-\theta)}\\\\$$
also, 140 degrees is in the 2nd quadrant. Sine is positive in the 2nd quad. so
$$\\sin140^0=sin(180-140)=sin40\\\\
sin40=cos50\\
so \\
sin140=cos50$$
This is the same as CPhill's answer.
Now we know that sin50=p=p/1
Draw a right angled triangle and label on of the acute angles as 50 degrees.
The opposite side is p
The hypotenuse is 1
So using pythagoras' Theorum the adjacent side must be $$\sqrt{1-p^2}$$
$$\\cos50=\frac{adj}{hyp}=\frac{\sqrt{1-p^2}}{1}\\\\
cos50^0=\sqrt{1-p^2}$$