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?

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- Feb 24th 2009, 07:54 AMmorelemonzanybody?re: solving trig. equations
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? - Feb 24th 2009, 08:00 AMReckoner
- Feb 24th 2009, 08:09 AMmorelemonzanybody?reply
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 - Feb 24th 2009, 08:21 AMReckoner
Yes. But if no restriction is given on $\displaystyle \phi,$ then you should account for the other solutions. In this case, the solutions are $\displaystyle \phi=60^\circ+360^\circ n,$ where $\displaystyle n$ is an integer (in other words, $\displaystyle 60^\circ$ and any angle that is coterminal with $\displaystyle 60^\circ$).

- Feb 24th 2009, 08:54 AMmorelemonzanybody?re:
hmmm!

that all makes sense to me.

but the answer they have given in the text is 22.5, 112.5 - Feb 24th 2009, 09:09 AMReckoner
Solutions are easy to verify:

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

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

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

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

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

$\displaystyle =\frac14+\frac34=1,$

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. - Feb 24th 2009, 09:11 AMmorelemonzanybody?re: restrictions
I didn't mention that the restriction is 0≤ø≤180

- Feb 24th 2009, 09:28 AMmorelemonzanybody?re: checking solutions
wow!

That is veeery cool!(Clapping)

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 - Feb 25th 2009, 04:22 AMekledessolve this equation
solve this equation

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