Question #21640

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Answer 1 · 2017-12-01T10:37:21

We can use orthogonal combinations of sines and cosines to prove that $\sin\theta=(3/5)$ .

You have the following:

$3\sin\theta+4\cos\theta=5$

Now construct an orthogonal relation by changing the $+$ sign to $-$ and switching the sine and cosine terms:

$4\sin\theta-3\cos\theta=a$

We have to find $a$ .

Square both equations:

$9\sin^2\theta+24\sin\theta\cos\theta+16\cos^2\theta=25$

$16\sin^2\theta-24\sin\theta\cos\theta+9\cos^2\theta=a^2$

And then add them up. Note that some terms cancel:

$25(\sin^2\theta+\cos^2\theta)=25+a^2$

But then, $\sin^2\theta+\cos^2\theta=1.$ And now we see $a$ has to equal zero!

So our original linear relations must be

$3\sin\theta+4\cos\theta=5$

$4\sin\theta-3\cos\theta=0$

Now just take three times the first equation plus four times the second making the cosines cancel:

$25\sin\theta=15$

$\sin\theta=(3/5)$

Answer 2 · 2017-12-01T11:22:10

$sintheta=3/5$

We are given $3sintheta+4costheta=5$

$3sintheta+4costheta$ can equal $R(costhetacosalpha+sinthetasinalpha)-=Rcos(theta-alpha)$

$R=sqrt(3^2+4^2)=sqrt(25)=5$

$3sintheta+4costheta=Rcosthetacosalpha+Rsinthetasinalpha$

$3=Rsinalpha$
$4=Rcosalpha$

$(Rsinalpha)/(Rcosalpha)=tanalpha=3/4$

$alpha=arctan(3/4)$ (We will leave $alpha$ as that to avoid any rounding errors)

$5cos(theta-arctan(3/4))=5$

$cos(theta-arctan(3/4))=1$

$theta-arctan(3/4)=arccos(1)=0$

$theta=arctan(3/4)$

$sintheta=sin(arctan(3/4))=3/5$