Prove $1<sin\,\theta + cos\,\theta \leq \sqrt 2$ for $0<\theta <90^o$

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In summary, the inequality $1<sin\,\theta + cos\,\theta \leq \sqrt 2$ means that for any angle $\theta$ between 0 and 90 degrees, the sum of the sine and cosine values must fall within a specific range. It is only true for angles between 0 and 90 degrees because the sine and cosine values fall within this range. This inequality can be proven using the Pythagorean identity and has significance in understanding the behavior of trigonometric functions within a specific range. It can also be applied to other trigonometric functions, but the range of angles for which it holds true may differ.
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Albert1
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$0<\theta <90^o$
using geometry to prove the following:
$1<sin\,\, \theta + cos\,\,\theta \leq \sqrt 2$
 
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Albert said:
$0<\theta <90^o$
using geometry to prove the following:
$1<sin\,\, \theta + cos\,\,\theta \leq \sqrt 2$
my soluion
$A,B,C,D$ are four points on a circle, with diameter $\overline {AC}= 1$
let :$\angle DAC=\theta , \angle CAB=\angle BCA=45^o $, then we have:
$\overline {AD}=cos\theta, \overline {CD}=sin\theta, \overline {AB}=\overline {BC}=\dfrac {\sqrt 2}{2}$
it is easy to see $cos\,\theta +sin\, \theta =\overline {AD}+\overline {CD}>\overline {AC}=1$
also by Ptolemy's theorem we have :
$\overline {AB}\times \overline {CD}+\overline {BC}\times\overline {AD}=\overline {AC}\times \overline {BD}=\overline {BD}\leq 1$
that is :$\dfrac {\sqrt 2}{2} sin\, \theta+\dfrac {\sqrt 2}{2} cos\, \theta\leq 1$
or $sin \theta+cos \theta \leq \sqrt 2$
 
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FAQ: Prove $1<sin\,\theta + cos\,\theta \leq \sqrt 2$ for $0<\theta <90^o$

1. What does the inequality $1

The inequality $1

2. Why is the inequality only true for angles between 0 and 90 degrees?

The inequality is only true for angles between 0 and 90 degrees because the sine and cosine values fall within this range. Beyond 90 degrees, the sine and cosine values become negative, and the inequality would no longer hold true.

3. How can we prove that $1

We can prove this inequality using the Pythagorean identity, which states that $sin^2\,\theta + cos^2\,\theta = 1$. By rearranging this identity, we get $sin\,\theta + cos\,\theta = \sqrt{1-sin^2\,\theta}$. Since $sin^2\,\theta$ is always less than or equal to 1 for angles between 0 and 90 degrees, we can substitute this value into the equation to get $sin\,\theta + cos\,\theta \leq \sqrt 2$. We can also see that the minimum value for $sin\,\theta + cos\,\theta$ is achieved when $sin^2\,\theta = 0$, which results in $sin\,\theta + cos\,\theta = 1$. Therefore, the inequality $1

4. What is the significance of the inequality $1

The inequality $1

5. Can this inequality be applied to other trigonometric functions?

Yes, this inequality can be applied to other trigonometric functions, such as tangent and secant, by using their respective identities and manipulating the equations. However, the range of angles for which the inequality holds true may differ for each function.

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