Question

if theta is a positive acute angle prove that sin theta + cos theta is greater than 1

Answer 1

$\theta$ is a positive acute angle

means, 0° < $\theta$ < 90° .

now, sin$\theta$ + cos$\theta$

= √2 [ sin$\theta$ × 1/√2 + cos$\theta$ × 1/√2 ]

= √2[sin$\theta$cos45° + cos$\theta$.sin45° ]

we know, sinA.cosB + cosA.sinB = sin(A + B)

so, [sin$\theta$cos45° + cos$\theta$.sin45° ] = sin($\theta$ + 45°)

then, sin$\theta$ + cos$\theta$ = √2sin($\theta$+45°)

here, 0° < $\theta$ < 90°

so, sin($\theta$+45°) > sin45° = 1/√2

so, √2sin($\theta$ + 45°) > √2 × 1/√2 = 1

hence, √2sin($\theta$ + 45°) > 1

hence proved

Answer 2

Answer:

Sinθ + Cosθ > 1  if θ is an acute angle

Step-by-step explanation:

if theta is a positive acute angle prove that sin theta + cos theta is greater than 1

Let say

Sinθ + Cosθ  = x

As θ is an acute angle so Sinθ & Cosθ are positive

so x is also positive

Squaring both  sides

=> (Sinθ + Cosθ)²  = x²

=> Sin²θ + Cos²θ + 2SinθCosθ = x²

Using Sin²θ + Cos²θ = 1  & Sin2θ = 2SinθCosθ

=> 1 + Sin2θ = x²

as θ is an acute angle so 2θ will lie either in 1st Quadrant or 2nd quadrant

0°<θ<90° => 0°<2θ<180° =>

=> Sin2θ is + ve  

=> 1 + (+ve)  = x²

=> x² > 1

=> x > 1

=> Sinθ + Cosθ > 1