The problem:

Prove that sec^2 theta + cos^2 theta can never be less than 2.

I can't figure out how to go about.

I have tried to take sec^ theta as 1/cos^2 theta and cos^2 theta as sin^2 theta, but got nowhere.

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- Nov 5th 2011, 09:53 PMDonTrig proof involving secant and cosine
The problem:

Prove that sec^2 theta + cos^2 theta can never be less than 2.

I can't figure out how to go about.

I have tried to take sec^ theta as 1/cos^2 theta and cos^2 theta as sin^2 theta, but got nowhere. - Nov 5th 2011, 10:32 PMsbhatnagarRe: Trig proof involving secant and cosine
Let us take, $\displaystyle f(\theta)=\sec^2{ \theta} + \cos^2{\theta}$.

Now, use your knowledge of calculus to find the minimum value of $\displaystyle f(\theta)$.

You will find that minimum value of $\displaystyle f(\theta)$ is 2.

Therefore:

$\displaystyle f(\theta) \geq 2$

$\displaystyle \sec^2{ \theta} + \cos^2{\theta} \geq 2$

http://www.mathhelpforum.com/math-he...fg24f85edf.gif - Nov 6th 2011, 03:26 AMDonRe: Trig proof involving secant and cosine
Thanks for the reply.

I am still hazy about Calculus, but I think that using algebraic identities like (a+b)^2 and (a-b)^2 this can be solved:

sec^2 theta =1/ cos^2 theta. Therefore, cos^2 theta can be taken as 'x' and the equation reduces to x^2 + 1/x^2.

x^2 + 1/x^2 - 2*x*1/x + 2*x*1/x.

(x^2 - 1/x^2)^2 + 2

This way, sec^2 theta + cos^2 theta can never be less than 2. Is this right? - Nov 6th 2011, 03:44 AMveileenRe: Trig proof involving secant and cosine
Yup it's correct, even if you didn't write it ordered or... well, logical.

$\displaystyle x^{2}+\frac{1}{x^{2}}> 2 \Leftrightarrow x^{2}+\frac{1}{x^{2}}-2>0 \Leftrightarrow x^{2}+\frac{1}{x^{2}}-2\cdot x^{2}\cdot\frac{1}{x^{2}}>0 \Leftrightarrow \left ( x-\frac{1}{x} \right )^{2}>0$