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Help me year 11 trig question (a) prove(i have done this part already) cos3a = 4cos^3 a - 3cosa (b) Show that cos40 is a root of the equation 8x^3 - 6x +1 = 0 I need help with part (b) only

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well you have part a) already done, so why not replace 'a' with 40 and see what happens cos3a = 4cos^3 a - 3cosa cos(3*40) = 4cos^3(40) - 3cos(40) cos(120) = 4cos^3(40) - 3cos(40) -1/2 = 4cos^3(40) - 3cos(40)
You could just plug in x=cos40

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all you need to do is find the cosine of 40 degrees and replace x in the equation. cos40=0.7660
Now multiply both sides of the last equation by 2 to get -1/2 = 4cos^3(40) - 3cos(40) 2*(-1/2) = 2*(4cos^3(40) - 3cos(40) ) -1 = 8cos^3(40) - 6cos(40) 0 = 8cos^3(40) - 6cos(40) + 1 8cos^3(40) - 6cos(40) + 1 = 0 so that confirms x = cos(40) is a root of 8x^3 - 6x +1 = 0
oh :) thank you so much. didnt have any clue what to do :)
@jim_thompson5910 nice ... @AonZ we can also find by replacing x with cos40(0.7660) but this is very long method. when you replace x in your equation with the stored value of .7660, the result is 0 indicating that cosine of 40 degrees is a root of the equation. and by this, jim had proven this!!! good job @jim_thompson5910
@jim_thompson5910 way is the best :) cause it uses trig which is the topic i am currently doing :D

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