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# if sin 50 = p then sin 140 = ?

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if sin 50 = p

then sin 140 = ?

Guest Oct 5, 2014

#2
+91436
+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
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#1
+80933
+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
+91436
+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
#3
+80933
0