How could I solve this trigonometric equation?

[Guest]

Sin 2 theta = cos 7pi/4

 

[arthur ohlsten]

sin 2@=cos 7pi/4

2pi=360 degrees
pi/4= 45 degrees
7pi/4=315 degrees
7pi/4=-45 degrees
cos 7pi/4 = 1/sqrt2

sin 2@= 2 sin@ cos@

sin2@=cos 7pi/4
2 sin@ cos@=1/sqrt2
both sides
4 sin^2@ cos^2@= 1/2
sin^2@ cos^2@=1/8

but cos^2@= 1-sin^2@
substitute
sin^2@[1-sin^2@]= 1/8
-sin^4@ +sin^2@=1/8
sin^4@ - sin^2@ + 1/8=0
let sin^2@=x
x^2-x+1/8=0
by quadratic equation
x=1+/- [1-1/2]^1/2
x=1+/- [1/2]^1/2
x=1+/-1/2 [2]^1/2
x=1+/-.707
x=1.707 OR x=.293
sin^2 @= 1.707 OR sin^2 @=.293

but sin can't be greater than 1 so sin^2 can't exceed 1 or sin^2@=1.707 is extraneous answer

sin^2@=.293
sin@=+/- sqrt .293
sin@=+/- .5413
@=+/- 33 degrees

please check for errors
Arthur

 

[soroban]

Hello, americo74!

\(\displaystyle \sin2\theta\:=\:\cos\left(\frac{7\pi}{4}\right)\)

First of all, we know that: \(\displaystyle \:\cos\left(\frac{7\pi}{4}\right)\:=\:\frac{\sqrt{2}}{2}\)

The equation becomes: \(\displaystyle \:\sin2\theta\:=\:\frac{\sqrt{2}}{2}\;\;\Rightarrow\;\;2\theta\:=\:\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)\)

The angles whose sine is \(\displaystyle \frac{\sqrt{2}}{2}\) are: \(\displaystyle \,\frac{\pi}{4}\,+\,2\pi n,\:\frac{3\pi}{4}\,+\,2\pi n\;\) for any integer \(\displaystyle n\)

So we have: \(\displaystyle \:2\theta\:=\:\left\{\frac{\pi}{4}\,+\,2\pi n,\:\frac{3\pi}{4}\,+\,2\pi n\right\}\)

Therefore: \(\displaystyle \L\:\theta\:=\:\left\{\frac{\pi}{8}\,+\,\pi n,\;\frac{3\pi}{8}\,+\,\pi n\right\}\;\;\) for any integer \(\displaystyle n.\)

 

[Guest]

one must always thank people like you...so THANK YOU

I appreciate your help...

 

[arthur ohlsten]

you are welcome
Arthur