On putting the value of sin( α +β +γ #160; ) #160; we get sinα #160; +sinβ #160; +sinγ #160; =sinα #160; ( 1 cosα #160; cosγ #160; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Dot Product

If 0<α ,β ,...

Question

If $0 < α, β, γ < π / 2$ , prove that $sin α + sin β + sin γ > sin (α + β + γ)$

Open in App

Solution

On putting the value of $sin (α + β + γ)$ we get
$sin α + sin β + sin γ = sin α (1 - cos α cos γ) + sin β (1 - cos γ cos α) + sin γ (1 - cos α cos β) + sin α sin β sin γ > 0$
$∵ α, β, γ$ all lie in 1st quadrant
$cos α cos β$ is +ve and less than $1$ so that $1 - cos α cos β$ is +ve and also $sin γ$ is +ive. Thus every term in LHS is +ive
$sin α sin β sin γ > sin (α + β + γ)$

Suggest Corrections

Similar questions

0 < α, β, γ < \frac{π}{2}

sin α + sin β + sin γ > sin (α + β + γ)

α, β, γ \in (0, \frac{π}{2})

\frac{sin (α + β + γ)}{sin α + sin β + sin γ}

α, β, γ ϵ (0, \frac{π}{2})

\frac{sin (α + β + γ)}{sin α + sin β + sin γ}

0 < α < β < γ < π / 2

tan α < \frac{sin α + sin β + sin γ}{cos α + cos β + cos γ} < tan γ

c o s α + c o s β + c o s γ = s i n α + s i n β + s i n γ = 0

c o s (β + γ) + c o s (γ + α) + c o s (α + β) = 0

s i n (β + γ) s i n (γ + α) + s i n (α + β) = 0

Join BYJU'S Learning Program