# sin and cos

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

$\displaystyle cosa\:=$

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

the second part of the question is

$\displaystyle sin2a\:=$

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

and from Pythagoras' theorem

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

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

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

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

$\displaystyle 90^o<a<180^o$

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..

$\displaystyle Cos(A)=-Cos(180-A)$
• 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...

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

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

thus;

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

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

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