Question

Use the fundamental identities and appropriate algebraic operations to simplify the following expression. (18 +tan x) (18-tan x)+ sec 2x Complete the following statement The lowest point on the graph of y = cos x, 0sxs2x, occurs when x-

Answer 1

Answer:

a) $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)=325$

b) The lowest point of $y=\cos \left(x\right)$, $0\leq x\leq 2\pi$ is when x = $\pi$

Step-by-step explanation:

a) To simplify the expression $\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)+\sec ^2\left(x\right)$ you must:

Apply Difference of Two Squares Formula: $\left(a+b\right)\left(a-b\right)=a^2-b^2$

$a=18,\:b=\tan \left(x\right)$

$\left(18+\tan \left(x\right)\right)\left(18-\tan \left(x\right)\right)=18^2-\tan ^2\left(x\right)=324-\tan ^2\left(x\right)$

$324-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Apply the Pythagorean Identity $1+\tan ^2\left(x\right)=\sec ^2\left(x\right)$

From the Pythagorean Identity, we know that $1=-\tan ^2\left(x\right)+\sec ^2\left(x\right)$

Therefore,

$324[-\tan ^2\left(x\right)+\sec ^2\left(x\right))]\\324[+1]\\325$

b) According with the below graph, the lowest point of $y=\cos \left(x\right)$, $0\leq x\leq 2\pi$ is when x = $\pi$