The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Some simple trig

Ok, I am going bonkers here. This should be really easy! But it's not making sense.

given π2<θ<π\frac{\pi}{2} < \theta < \pi

and cos⁡θ=−15\cos\theta=-\frac{1}{5}

find
cos⁡2θ\cos 2\theta
sin⁡θ\sin\theta
sin⁡2θ\sin 2\theta

using the double angle formulae, I get

cos⁡2θ=−2325\cos 2\theta=-\frac{23}{25}

sin⁡θ=256\sin\theta=\frac{2}{5}\sqrt 6

sin⁡2θ=−42526\sin 2\theta=-\frac{4}{25}\sqrt{26}

But these don't seem to convert back to the correct angles?

I seem to have confused myself. Can someone check my answers please?
(edited 13 years ago)
I presume the last one is a typo, and that's meant to be root(6), not root(26).

Answers seem fine to me.
Original post by ghostwalker
I presume the last one is a typo, and that's meant to be root(6), not root(26).

Answers seem fine to me.


ah... yes indeed it is. 6\sqrt{6} not 26\sqrt{26}

But they don't seem to produce the right angles, do they? (when arcsin-ed etc)

Maybe they do. Can't think now. It's too late. I'll have another look tomorrow.

Thanks anyway.
Original post by Plato's Trousers
...


The way I'd check would be to find θ\theta from your initial value of cos⁡θ\cos\theta and then evaluate the three functions; don't attempt to do inverse trig. on them.

If you really want to take inverse trig functions of the three, then you need to be careful of which quadrant they are refering to.
Original post by ghostwalker
The way I'd check would be to find θ\theta from your initial value of cos⁡θ\cos\theta and then evaluate the three functions; don't attempt to do inverse trig. on them.

If you really want to take inverse trig functions of the three, then you need to be careful of which quadrant they are refering to.


yes, it actually all works out fine. Not quite sure why I couldn't do this yesterday. It's very elementary stuff.

Thanks

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you