trigonometry (3)

• Oct 7th 2009, 12:37 AM
thereddevils
trigonometry (3)
If $\displaystyle 0<A<\frac{\pi}{2}$ , and $\displaystyle 0<B<\frac{\pi}{2}$ , prove that

(1)$\displaystyle \frac{1}{\cos^2 A}+\frac{1}{\sin^2 A\sin^2 B\cos^2 B}\geq 9$

(2) Prove that $\displaystyle sin^2 A + sin^2 B > sin A + sin B + sin Asin B -1$
• Oct 7th 2009, 08:50 AM
red_dog
1) We have

$\displaystyle \frac{1}{\sin^2A\sin^2B\cos^2B}=\frac{\sin^2B+\cos ^2B}{\sin^2A\sin^2B\cos^2B}=\frac{1}{\sin^2A\cos^2 B}+\frac{1}{\sin^2A\sin^2B}$

Then we have to prove that

$\displaystyle \frac{1}{\cos^2A}+\frac{1}{\sin^2A\cos^2B}+\frac{1 }{\sin^2A\sin^2B}\geq 9$

Use the inequality $\displaystyle (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\r ight)\geq 9$

for $\displaystyle a=\cos^2A, \ b=sin^2A\cos^2B, \ c=\sin^2A\sin^2B$

using that

$\displaystyle a+b+c=\cos^2A+\sin^2A\cos^2B+\sin^2A\sin^2B=$

$\displaystyle =\cos^2A+\sin^2A(\sin^2B+\cos^2B)=\cos^2A+\sin^2A= 1$
• Oct 8th 2009, 03:38 AM
Quote:

Originally Posted by thereddevils
If $\displaystyle 0<A<\frac{\pi}{2}$ , and $\displaystyle 0<B<\frac{\pi}{2}$ , prove that

(1)$\displaystyle \frac{1}{\cos^2 A}+\frac{1}{\sin^2 A \sin^2 B \cos^2 B} \geq 9$

(2) Prove that $\displaystyle sin^2 A + sin^2 B > sin A + sin B + sin Asin B -1$

For (1) $\displaystyle \sec^2 A + \text{cosec}^2 A \, \text{cosec}^2 B \, \sec^2 B$

$\displaystyle =(1+\tan^2 A)+(1+\cot^2 A)(1+\cot^2 B)(1+tan^2 B)$

$\displaystyle =(1+\tan^2 A)+(1+\cot^2 A)(2+\tan^2 B+\cot^2 B)$

Consider $\displaystyle (\tan B-\cot B)^2\geq 0$

$\displaystyle \tan^2 B+\cot^2 B \geq 2$

Then $\displaystyle (\tan A-2\cot A)^2\geq 0$

$\displaystyle \tan^2 A+4\cot^2 A\geq 4$

continuing

$\displaystyle =(1+\tan^2 A)+(1+\cot^2 A)(4)$

$\displaystyle =5+\tan^2 A+4\cot^2 A$

$\displaystyle =9$

this inequality is at least 9 . Hence , proved .