A simple equation involving trigonometry If cos⁡θ+sin⁡θ+tan⁡θ=2+1 then what is the maximum value of sin⁡θ,sin2θ?

If you use the tangent half-angle substitution, you end with t4−2t2+4t+11−t4=2+1 and one of the roots is t=2−1. You also have two complex roots and another real which is really messy.

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petrusrexcs

Answered question

2022-01-17

A simple equation involving trigonometry
If

\cos θ + \sin θ + \tan θ = \sqrt{2} + 1

then what is the maximum value of

\sin θ, \sin^{2} θ

Answer & Explanation

soanooooo40

Beginner2022-01-18Added 35 answers

Let

a = \sin θ, b = \cos θ

. Then

a^{2} + b^{2} = 1

and

a + b + \frac{a}{b} = \sqrt{2} + 1

Notice that:

a + b \leq | a + b | \leq \sqrt{2 (a^{2} + b^{2})} = \sqrt{2}

and thus we deduce that

\frac{a}{b} \geq 1

. So a,b have the same sign. Since we want the maximum of a, we are only interested in the case when a and b are positive. In this case, using AM-GM:

\sqrt{2} + 1 = a + b + \frac{1}{\sqrt{2}} \frac{a}{b} + (1 - \frac{1}{\sqrt{2}}) \frac{a}{b} \geq 3 \sqrt{3} {\frac{a^{2}}{\sqrt{2}}} + 1 - \frac{1}{\sqrt{2}}

and from here, it's easy to find that

a \leq \frac{1}{\sqrt{2}}

temzej9

Beginner2022-01-19Added 30 answers

If you use the tangent half-angle substitution, you end with

\frac{t^{4} - 2 t^{2} + 4 t + 1}{1 - t^{4}} = \sqrt{2} + 1

and one of the roots is

t = \sqrt{2} - 1

. You also have two complex roots and another real which is really messy.

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