2arctan(1/2)+arccos(-3/5)=x how to solve?

[Wi_N]

Homework Statement

I need their sum, i know its pi but i have no idea how to get that.

Homework Equations

The Attempt at a Solution

no idea where to even begin solve it algebraic or analytical. convert them into tan and cos I am clueless.
any guidance would be appreciated.

 

[Dick]

Homework Statement

I need their sum, i know its pi but i have no idea how to get that.

Homework Equations

The Attempt at a Solution

no idea where to even begin solve it algebraic or analytical. convert them into tan and cos I am clueless.
any guidance would be appreciated.

I would start by taking e.g. $\cos$ of both sides.

 

[willem2]

You can use the trigonometric addition formulas.
start with tan(2 arctan (1/2)). use tan(2x) = 2 tan(x)/(1-(tan(x))^2)

Then you compute the cosine of the whole expression with the addition formula for cos(a+b).

you'll have to compute terms like cos (atn(a)). If you draw a right triangle with one angle equal to atn(a) it should be easy to see what cos(atn(a)) is.

 

[ehild]

As the first term is twice the angle α = arctan(1/2), assume that arcos(-3/5)=2β. Use the double angle forrmula to get cos²β. Following @willem2's hint, you get how tan²β and cos²β are related, so you get tanβ. How is it related to tan α?

 

[fresh_42]

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