Question 3o show all workFind the sum of2i 3(a 1+an

Mathematics

1 answer:

xeze [42]3 years ago

8 0

$n=8\\
a_1=2\cdot3+3=9\\
a_n=a_8=2\cdot10+3=23\\\\
\displaystyle
\sum\limits_{i=3}^{10}2i+3=S_8=\dfrac{8}{2}\cdot(9+23)=4\cdot32=128$

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To prove:

$\cos 3\theta=3\cos^3 \theta-3\sin^2 \theta\cos \theta$

Identities used:

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Taking the LHS:

$\Rightarrow \cos 3\theta=\cos (\theta +2\theta )$

Using identity 1:

$\Rightarrow \cos (\theta +2\theta )=\cos \theta \cos 2\theta-\sin \theta \sin 2\theta$

Using identities 2 and 3:

$\Rightarrow \cos \theta (\cos ^2\theta-\sin^2 \theta )-\sin \theta (2\sin \theta \cos \theta )\\\\\Rightarrow \cos^3\theta -\sin^2\theta \cos \theta -2\sin^2 \theta \cos \theta \\\\\Rightarrow \cos^3\theta -3\sin^2\theta \cos \theta$

Using identity 4:

$\Rightarrow \cos^3\theta -3(1-\cos^2\theta) \cos \theta\\\\\Rightarrow \cos^3\theta-3\cos \theta+3\cos^3\theta\\\\\Rightarrow 4\cos^3\theta -3\cos \theta$