Question

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

Answer 1

Answer:

Step-by-step explanation:

We know that,

sin(A+B) =sin(A)cos(B) +cos(A)sin(B)

Here,

we have

$\theta \:is\: acute ,i.e.,\theta \in(0,90\degree)$

Then,

$sin(\theta)+cos(\theta) = \sqrt{2}(\frac{1}{ \sqrt{2}} sin(\theta)+\frac{1}{ \sqrt{2}} cos(\theta))$

$sin(\theta)+cos(\theta) = \sqrt{2}(cos(45\degree) sin(\theta)+sin(45\degree)} cos(\theta))$

$\sqrt{2}(sin(45\degree+\theta))$

Now,

since sine function is increasing from 0 to 90degree , and decreasing from 90 degree to 180degree .

$f(x)=\sqrt{2}(sin(45\degree+\theta) \\f(x)_{min} = f(0^+) >1. \\ \\ f(x)_{max} =f(45\degree)=  \sqrt{2}(sin(45\degree+\theta) \\ \\ \sqrt{2}(sin(90\degree) = \sqrt{2}.$

Hence, f(x) is greater than 1.

$=> \sin(\theta) +cos(\theta) >1 , \forall \theta \in(0,90\degree)$