1. ## 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

$\displaystyle sin2a\:=$

$\displaystyle cos2a\:=$

2. $\displaystyle SinA=\frac{opposite}{hypotenuse}$

and from Pythagoras' theorem

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

3. 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 $\displaystyle -\frac{3}{5}$

4. 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..

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

5. here is a sketch

6. 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!

7. That's the idea,
you find the answers of the "double angle" situation,
using your results with the "single angle".

Good work!