#2**+10 **

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}$$

Melody
Oct 5, 2014

#1**+5 **

Using an additive identity (and assuming degrees), we have

sin 140 =

sin(90 + 50) =

sin90cos50 + sin50cos90 =

1*cos50 + p*0 =

cos50

CPhill
Oct 5, 2014

#2**+10 **

Best Answer

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}$$

Melody
Oct 5, 2014