# sin and cos

• Jan 24th 2010, 07:55 AM
davidman
sin and cos
When $\frac{\pi}{2}, $sina=\frac{4}{5}$

$cosa\:=$

not quite sure how to go from sin to cos... not quite sure what to do...

the second part of the question is

$sin2a\:=$

$cos2a\:=$
• Jan 24th 2010, 08:04 AM
$SinA=\frac{opposite}{hypotenuse}$

and from Pythagoras' theorem

$adjacent=\sqrt{(hypotenuse)^2-(opposite)^2}$
• Jan 24th 2010, 07:58 PM
davidman
Ok,

$adjacent=\sqrt{(hypotenuse)^2-(opposite)^2}=3$

$cosA=\frac{adjacent}{hypotenuse}=\frac{3}{5}$

so far so good, I guess... but the book says the answer is $-\frac{3}{5}$
• Jan 25th 2010, 02:05 AM
This is because of the range given,

$90^o

Cosine locates the x co-ordinate of a point on the circle circumference.
Once you have discovered the "distance" the point is from the origin, you simply invert it in this case as the point lies to the left of the y-axis..

$Cos(A)=-Cos(180-A)$
• Jan 25th 2010, 02:40 AM
here is a sketch
• Jan 25th 2010, 06:01 AM
davidman
Ok, thanks... googled around a bit and found doubleangle formulae...

$sin2a=2sin\:a\:cos\:a\:\:,\:\:cos2a=cos^2a-sin^2a$

and I'm pretty sure this made an appearance in my earlier thread now that I think of it...

thus;

$sin2a=2sin\:a\:cos\:a=2\times{\frac{4}{5}}\times{-\frac{3}{5}}=-\frac{24}{25}$

$cos2a=cos^2a-sin^2a=(-\frac{3}{5})^2-(\frac{4}{5})^2=-\frac{7}{25}$

Many thanks! (Rock)
• Jan 25th 2010, 07:52 AM