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Run hard@thehsc · Mar 21, 2022

for (iv) I am getting cos theta = 0 and cos theta = 1/quadruple root 32
I am not sure why, but the second solution seems off - can someone help me with this here

cossine · Mar 21, 2022

cos(theta)(16cos^4 theta -1) = 0

cos^4 theta = 1/16

=> cos(theta) =1/2 Since 1/16 = 2^-4 and (2^-4)^1/4 = 2^-1

5uckerberg · Mar 21, 2022

Run hard@thehsc said:
View attachment 35242
for (iv) I am getting cos theta = 0 and cos theta = 1/quadruple root 32
I am not sure why, but the second solution seems off - can someone help me with this here

The hint is already given. Well, $16\cos^{5}\theta=\cos{5\theta}+5\cos{3\theta}+10\cos{\theta}$

Then part iv is just simplified into $\left(\cos{\theta}\right)\left(16\cos^{4}{\theta}-1\right)=0$ from $16\cos^{5}\theta-\cos{\theta}$ using part ii.

Thus, $\cos{\theta}=0, \left(16\cos^{4}{\theta}-1\right)=0$

$\cos{\theta}=\frac{1}{2}, 0$

There is where things get really entertaining because $\theta=\cos^{-1}0, \cos^{-1}\left(\frac{1}{2}\right)$

These two give these values of $\theta = \frac{\pi}{2}, \frac{\pi}{3}$

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Run hard@thehsc

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