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?

.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 to start.







Now, you know that

Originally Posted by Quacky

<--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.







Now, you know that

Of course it is! Thank you so much.