Question

prove the following
${8 \cos(40) }^{3} = (6 \cos(40) - 1)$
​

Answer 1

Answer:

$\huge\mathtt\red{= 1}$

Step-by-step explanation:

we know  

cos30=4cos30-3cos0 

So, 2cos(30)=8cos30−6cos0  

∴2cos3(20∘)-6cos(20∘) 

=2cos(×20∘)  

2cos(3×9π) 

2cos(3π)  

2×1/2$$

=1