3U Trig Identity Question Help (1 Viewer)

KaoticKreature · Sep 17, 2016

Hey guys,
Can you guys please help with this identity? Cambridge doesn't provide proofs. This is question 13m from chapter 2A of the Cambridge Year 12 Textbook. Thanks guys.
https://i.gyazo.com/934bde58b5ffb5259c2aed97eac9a52f.png

hennychu12 · Sep 17, 2016

At that point you should just shoot yourself

InteGrand · Sep 17, 2016

KaoticKreature said:
Hey guys,
Can you guys please help with this identity? Cambridge doesn't provide proofs. This is question 13m from chapter 2A of the Cambridge Year 12 Textbook. Thanks guys.
https://i.gyazo.com/934bde58b5ffb5259c2aed97eac9a52f.png

$\noindent For simplicity, let $s \equiv \sin \alpha$ and $c \equiv \cos \alpha$. Call the L.H.S. of the identity we are asked to prove $L$ and the R.H.S. $R$. Then$

$\begin{align*}L &= \frac{2sc + 2c^2 -1}{2c + s -2\left(c^3 + s^3 \right)}\quad (\text{double angle formulas in numerator}).\end{align*}$

$\noindent The denominator here is$

$\begin{align*}2c + s -2c^3 -2s^3 &= 2c \left(1-c^2\right) +s \left(1-2s^2\right) \\ &= 2cs^2 +s\left(-1 + 2c^2\right) \quad (\text{since }c^2 + s^2 = 1) \\ &= s\left(2sc -1 + 2c^2\right),\end{align*}$

$\noindent which is simply $s$ times the numerator of $L$. This implies that $L = \frac{1}{s}= R$, as required.$

KaoticKreature · Sep 17, 2016

Thanks man! That's really helpful. Out of curiosity, how did you know to change the cos2x to 2cos^2x-1 rather than for example cos^2x - sin^2x??? How did you know which one to use? Thanks!

InteGrand · Sep 17, 2016

KaoticKreature said:
Thanks man! That's really helpful. Out of curiosity, how did you know to change the cos2x to 2cos^2x-1 rather than for example cos^2x - sin^2x??? How did you know which one to use? Thanks!

$\noindent It doesn't matter, you could use any form of the double angle formula for it. Whichever one you used, your aim would be to show the denominator is $s$ times the numerator. You'd be able to still do this by tweaking what I showed the denominator is with the use of the relationship $c^2 + s^2 = 1$.$

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3U Trig Identity Question Help (1 Viewer)

KaoticKreature

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hennychu12

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InteGrand

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KaoticKreature

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InteGrand

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