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**CSG18** · Mar 13th 2010, 01:14 PM

Based on the following information from a past paper:

Since p = sin40 and q = cos110.

and therefore sin140 = p and cos70 = - q

One of the questions was:find an expression for cos140

In the mark scheme the answer was:

cos^2(X) + sin^2(X) = 1 (I understand the importance of using the trig identity)

followed by: cos(140) = √(1 - sin^2(X)) (I got this far)

this is the bit which I do not understand which was in the mark scheme:

cos(140) = √(1 - p^2)

How does p = sin x ?

all the information that it gives is that p = sin40, so how can that be applied in general?

**Quacky** · Mar 13th 2010, 01:46 PM

$cos^2(x)+sin^2(x)=1$ .This is the identity, it is not being applied to the example. The mark scheme doesn't actually say this, it would seem, but you should say that $x=140$ to start.

$cos^2(140)+sin^2(140)=1$

$cos^2(140)=1-sin^2(140)$

$cos(140)=\sqrt{1-sin^2(140)}$

Now, you know that $sin(140)=p$

$<br /> cos(140)=\sqrt{1-(p)^2}<br />$

**CSG18** · Mar 13th 2010, 01:50 PM

Originally Posted by Quacky

$cos^2(x)+sin^2(x)=1$ <--this is the identity, it is not being applied to the example. The mark scheme doesn't actually say this, it wouls seem, but you should say that x=140 to start.

$cos^2(140)+sin^2(140)=1$

$cos^2(140)=1-sin^2(140)$

$cos(140)=\sqrt{1-sin^2(140)}$

Now, you know that $sin(140)=p$

$<br /> cos(140)=\sqrt{1-(p)^2}<br />$

Of course it is! Thank you so much.

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