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Trigonometric Ratios of Multiples of an Angle

If α=sin -1√3...

Question

If α = sin^{-1 then}

A

∝ > β

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B

∝ = β

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C

∝ < β

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D

∝ + β = 2 π

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Solution

The correct option is C
∝ < β

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Similar questions

α = {sin}^{- 1} \frac{\sqrt{3}}{2} + {sin}^{- 1} \frac{1}{3}

β = {cos}^{- 1} \frac{\sqrt{3}}{2} + {cos}^{- 1} \frac{1}{3}

α, β, γ

are three angles given by

α = 2 t a n^{- 1} (\sqrt{2} - 1), β = 3 s i n^{- 1} \frac{1}{\sqrt{2}} + s i n^{- 1} (\frac{- 1}{2}) a n d γ = c o s^{- 1} (\frac{1}{3}),

α, β

γ

are three angles given by

α = 2 {tan}^{- 1} (\sqrt{2} - 1), β = 3 {sin}^{- 1} \frac{1}{\sqrt{2}} + {sin}^{- 1} (- \frac{1}{2})

γ = {cos}^{- 1} \frac{1}{3} .

Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

0 \leq x \leq \frac{1}{2}

then the value of

t a n [{s i n}^{- 1} {\frac{x}{\sqrt{2}} + \frac{\sqrt{1 - x^{2}}}{\sqrt{2}}} - {s i n}^{- 1} x]

α = {s i n}^{- 1} \frac{4}{5} + {s i n}^{- 1} \frac{1}{3}

β = {c o s}^{- 1} \frac{4}{5} + {c o s}^{- 1} \frac{1}{3}

α

β

are acute angles such that

c o s^{2} α + c o s^{2} β = 3 / 2

α

β

α + β

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