Trig Identity Question (1 Viewer)

Speed6 · Jul 27, 2015

Trying all sorts of ways to get LHS to = RHS, came close, but never got it. Could someone possibly help with the proof?

$\frac{sin2A+1}{cos2A}=\frac{cosA+sinA}{cosA-sinA}$

rand_althor · Jul 27, 2015

Speed6 · Jul 27, 2015

Thanks, nice setting out too- I like it

integral95 · Jul 27, 2015

now to get LHS to RHS

$LHS = \frac{sin2A+1}{cos2A} = \frac{sin^2A+2sinAcosA + cos^2A}{cos^2A-sin^2A} = \frac{(sinA+cosA)^2}{(cosA+sinA)(cosA-sinA)} = \\ \\ \\ \frac{cosA+sinA}{cosA-sinA} = RHS$

Speed6 · Jul 27, 2015

integral95 said:
now to get LHS to RHS

$LHS = \frac{sin2A+1}{cos2A} = \frac{sin^2A+2sinAcosA + cos^2A}{cos^2A-sin^2A} = \frac{(sinA+cosA)^2}{(cosA+sinA)(cosA-sinA)} = \\ \\ \\ \frac{cosA+sinA}{cosA-sinA} = RHS$

Just to clarify, $LHS = \frac{sin2A+1}{cos2A} = \frac{sin^2A+2sinAcosA+ cos^2A}{cos^2A-sin^2A}$

the numerator of $sin^2A+cos^2A$ addition was achieved by the reciprocal of $cos^2A-sin^2A$ ?

Thanks for that

InteGrand · Jul 27, 2015

Speed6 said:
Just to clarify, $LHS = \frac{sin2A+1}{cos2A} = \frac{sin^2A+2sinAcosA+ cos^2A}{cos^2A-sin^2A}$

the numerator of $sin^2A+cos^2A$ addition was achieved by the reciprocal of $cos^2A-sin^2A$ ?

Thanks for that

$The 1 became $\sin^2 A + \cos^2 A$ because $1=\sin^2 A + \cos^2 A$ (Pythagorean trig. identity).$

Speed6 · Jul 27, 2015

InteGrand said:
$The 1 became $\sin^2 A + \cos^2 A$ because $1=\sin^2 A + \cos^2 A$ (Pythagorean trig. identity).$

Makes much more sense

integral95 · Jul 27, 2015

Speed6 said:
Just to clarify, $LHS = \frac{sin2A+1}{cos2A} = \frac{sin^2A+2sinAcosA+ cos^2A}{cos^2A-sin^2A}$

the numerator of $sin^2A+cos^2A$ addition was achieved by the reciprocal of $cos^2A-sin^2A$ ?

Thanks for that

I used
$sin^2A+cos^2A = 1$

Speed6 · Jul 27, 2015

The next question I have:

Use the 't' method to prove that:

$tan\theta = \frac{sin2\theta +sin\theta }{1+cos\theta + cos2\theta }$

rand_althor · Jul 27, 2015

Speed6 · Jul 27, 2015

Thanks for the proof

$Q.E.D$

braintic · Jul 28, 2015

rand_althor said:
$\begin{align*}\textup{Let }\tan{\theta}&=t \\\textup{RHS}&=\frac{\sin{2\theta}+\sin{\theta}}{1+\cos{2\theta}+\cos{\theta}} \\&=\frac{\dfrac{2t}{1+t^2}+\dfrac{t}{\sqrt{1+t^2}}}{1+\dfrac{1}{\sqrt{1+t^2}}+\dfrac{1-t^2}{1+t^2}} \\&=\frac{\dfrac{2t+t\sqrt{1+t^2}}{1+t^2}}{\dfrac{1+t^2+\sqrt{1+t^2}+1-t^2}{1+t^2}} \\&=\frac{t(2+\sqrt{1+t^2})}{2+\sqrt{1+t^2}} \\&=t \\&=\tan{\theta} \\ &=\textup{LHS}\end{align*}$

That only proves the result for acute theta.

Correction: For 1st and 4th quadrant angles only.

InteGrand · Jul 28, 2015

Speed6 said:
The next question I have:

Use the 't' method to prove that:

$tan\theta = \frac{sin2\theta +sin\theta }{1+cos\theta + cos2\theta }$

$We don't even need to use $t$-method to prove this; just use the double angle formulas $\sin 2\theta = 2\sin \theta \cos \theta$ (on the numerator) and $\cos 2\theta = 2\cos^2 \theta -1$ (on the denominator). This avoids needing to consider $\theta$ in various quadrants, so is better in my opinion than the $t$-method they are saying to use (they're probably just telling you to use it to get practice with it).$

Trig Identity Question (1 Viewer)

Speed6

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