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#