re: solving trig. equations

**morelemonzanybody?** · February 24th 2009, 07:54 AM

Hi:
sin30cosø+cos30sinø=1
sin(30+ø)=1
sin30+sinø=1
sinø=1-sin30, (1-1/2)
ø=sin^-1(1/2)

Can anybody tell me where I have gone wrong with this please?

**Reckoner** · February 24th 2009, 08:00 AM

Originally Posted by morelemonzanybody?

Hi:
sin30cosø+cos30sinø=1
sin(30+ø)=1
sin30+sinø=1 <--- Right here. In general, $\color{red}\sin(u+v)\neq\sin u+\sin v$
sinø=1-sin30, (1-1/2)
ø=sin^-1(1/2)

Can anybody tell me where I have gone wrong with this please?

After step 2, take the arcsine of both sides and then solve for $\phi$ (keeping in mind that there will be infinitely many solutions since the arcsine function will only give the one angle that is in its range).

**morelemonzanybody?** · February 24th 2009, 08:09 AM

ah! I think I am making the same mistake again.
so, should i:
sin(30+ø)=1
30+ø=sin^-1(1)
ø=sin^-1(1)-30=60

**Reckoner** · February 24th 2009, 08:21 AM

Originally Posted by morelemonzanybody?

ah! I think I am making the same mistake again.
so, should i:
sin(30+ø)=1
30+ø=sin^-1(1)
ø=sin^-1(1)-30=60

Yes. But if no restriction is given on $\phi,$ then you should account for the other solutions. In this case, the solutions are $\phi=60^\circ+360^\circ n,$ where $n$ is an integer (in other words, $60^\circ$ and any angle that is coterminal with $60^\circ$ ).

**morelemonzanybody?** · February 24th 2009, 08:54 AM

hmmm!
that all makes sense to me.
but the answer they have given in the text is 22.5, 112.5

**Reckoner** · February 24th 2009, 09:09 AM

Originally Posted by morelemonzanybody?

hmmm!
that all makes sense to me.
but the answer they have given in the text is 22.5, 112.5

Solutions are easy to verify:

$\sin30^\circ\cos\phi+\cos30^\circ\sin\phi=1$

Substituting $\phi=22.5^\circ,$ my calculator gives about 0.7934 for the left-hand side, which is certainly far from 1. With $\phi=112.5^\circ,$ I get $0.6088,$ which again does not equal 1. So these solutions do not work.

On the other hand, substituting $\phi=60^\circ$ gives

$\sin30^\circ\cos60^\circ+\cos30^\circ\sin60^\circ$

$=\left(\frac12\right)\left(\frac12\right)+\left(\f rac{\sqrt3}2\right)\left(\frac{\sqrt3}2\right)$

$=\frac14+\frac34=1,$

so our solution is correct.

Make sure you copied the problem down correctly, and that you were looking in the right section for the solution. It is entirely possible that the book has an error; such mistakes can occur quite often, unfortunately.

**morelemonzanybody?** · February 24th 2009, 09:11 AM

I didn't mention that the restriction is 0≤ø≤180

**morelemonzanybody?** · February 24th 2009, 09:28 AM

wow!
That is veeery cool!

I haven't thought to check solutions that way. That could really offer peace of mind about whether one has the right answer, or not.
Actually, this text has had quite a number of errors reported. Unfortunate, but I am glad I can now check myself. It is a distance learning book and not for general publication, so I guess there are few people viewing it apart from those studying it. It must be hard to write and type up so much maths.
Thanx Reckoner - onto the next problem

**ekledes** · February 25th 2009, 04:22 AM

solve this equation

sin 30 cos x + cos 30 sinx = 1 where x is real number

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