Question

Exam. 16.09.2012)
85. The minimum value of sin20 +
cos20 + sec20 + cosec20 + tan20
+ cot20 is
(2) 3 nie
3
(1) 1
(3) 5
(4) 7
92. TL​

Answer 1

Answer:

(4), The minimum value is 7.

Step-by-step explanation:

This is essentially an exercise in trigonometric identities.

First we notice that we can apply the pythogorean identity for sine and cosine:

$\sin^2 x + \cos^2 x = 1$

Applying it gives

$1 + \sec^2 \theta+ \csc^2 \theta + \tan^2 \theta + cot^2 \theta.$

Then we notice that dividing the pythogorean identity by $\cos^2 \theta$ gives

$tan^2 \theta + 1 = \frac{1}{\cos^2\theta} = \sec^2 \theta$.

Applying this gives us

$2\sec^2 \theta + csc^2 \theta + cot^2 \theta$.

The pythogorean identity can also be used to create an identity with cosecant and cotangent by dividing by $\sin^2 \theta$. This gives

$1 + \cot^2 \theta = \csc^2 \theta$.

Rearranging it gives

$\cot^2\theta = \csc^2 \theta -1$,

which we can apply to our expression to get

$2\sec^2\theta + 2\csc^2\theta - 1 = 2(\sec^2\theta + \csc^2\theta) -1$.

So now, we only need to minimize $sec^2\theta$ and $\csc^2\theta$. We know that

$sec^2\theta = \frac{1}{\cos^2\theta}$   and that   $\csc^2\theta = \frac{1}{sin^2\theta}$.

As secant and cosecant are minimum whenever cos and sine are at their maximum, then we now need to find the maximal values for sine and cosine.

This is when $\theta = \frac{\pi}{4} + 2n, \, n \in \mathbb{Z}$.

Hence, inputing $\theta = \frac{\pi}{4}$ into the expression gives

$\quad 2\left( \frac{1}{\cos^2 \frac{\pi}{4}} + \frac{1}{\sin^2 \frac{\pi}{4}} \right) -1 \\= 2\left( \frac{1}{\left(\frac{1}{2}} \right)} + \frac{1}{\left(\frac{1}{2}} \right)} \right) -1\\= 2(2 + 2 ) -1 = 8-1 = 7.$

Wich is over answer.