TRIG HW HELP...identities?

tan3x

= tan(x + 2x)

= (tanx + tan2x) / (1 - tanx tan2x)

= [tanx + 2tanx/(1 - tan^2 x)] / [1 - tanx * 2tanx/(1 - tan^2 x)]

= [(tanx - tan^3 x + 2tanx) / (1 - tan^2 x)] / (1 - tan^2 x - 2tan^2 x) / (1 - tan^2 x)]

= (3tanx - tan^3 x) / (1 - 3tan^2 x).

ok, it is the type you do it with calculus. permit's say a= first huge form b=2d numbers a + b = 20 a^2+b^2 = some huge form now you remedy the 1st equation for a or b, it is not considerable a = 20 - b and you replace (20-b)^2+b^2= some huge form now you opperate (b^2-40*b+4 hundred)+b^2 = 2*b^2-40*b+4 hundred you could now do 2 issues, you could graph or divide, in case you graph, then graph the functionality f(b)=2*b^2-40*b+4 hundred now in case you derive you get f'(b)= 4*b-40 and you're making it equatl to 0 4*b-40 = 0 and you remedy for b b = 40/4 or b = 10 and you could now get a a = 20-b a = 20-10 = 10 so then you fairly get that a = 10 and b = 10 that provides you with the minimum you like and you do the comparable with huge form 2 a + b = 30 a^2+b^2 = some huge form now you remedy the 1st equation for a or b, it is not considerable a = 30 - b and you replace (30-b)^2+b^2= some huge form now you opperate (b^2-60*b+900)+b^2 = 2*b^2-60*b+900 you could now do 2 issues, you could graph or divide, in case you graph, then graph the functionality f(b)=2*b^2-60*b+900 now in case you derive you get f'(b)= 4*b-60 and you're making it equivalent to 0 4*b-60 = 0 and you remedy for b b = 60/4 or b = 15 and you could now get a a = 30-b a = 30-15 = 15 so then you fairly get that a = 15 and b = 15 that provides you with the minimum you like wish this facilitates. pd. equivalent to 0 once you derive because of the fact meaning that the slope is going to be 0 and the slope will become 0 whilst it is maximum or minimum values.

tan3x = tan(x + 2x)

= (tanx + tan2x) / (1 - tanx tan2x)

= [tanx + 2tanx/(1 - tan^2 x)] / [1 - tanx * 2tanx/(1 - tan^2 x)]

= [(tanx - tan^3 x + 2tanx) / (1 - tan^2 x)] / (1 - tan^2 x - 2tan^2 x) / (1 - tan^2 x)]

= (3tanx - tan^3 x) / (1 - 3tan^2 x)

tan(3x)=tan(x+2x)

Tan(a+b)=(Tana+tanb)/(1-tanatanb)

So, tan(x+2x)=(tanx+tan2x)/(1-tanxtan2x)

Write tan2x=2tanx/(1-(tan(x))^2) and simplify