Thread: More expressions

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?

$\displaystyle cos^2(x)+sin^2(x)=1$.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 $\displaystyle x=140$ to start.

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

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

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

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

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

Originally Posted by Quacky

$\displaystyle cos^2(x)+sin^2(x)=1$<--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.

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

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

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

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

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

Of course it is! Thank you so much.