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2 years ago

Simplify - root (9+y)^2 if y <-10

Mathematics

1 answer:

nevsk [136]2 years ago

6 0

Answer:

− √ ( 9 + y ) 2 > 1

Explanation:

− √ ( 9 + y ) 2 , where y < − 10

Simplify the expression,

− √ ( 9 + y ) 2 = − ( 9 + y ) = − 9 − y

When y < − 10 , − y > 10

Subtract 9 from both sides,

− 9 − y > 1 Hence,

− √ ( 9 + y ) 2 > 1

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find the values of the six trigonometric functions for angle theta in standard position if a point with the coordinates (1, -8)

frutty [35]

Answer:

cosФ = $\frac{1}{\sqrt{65}}$ , sinФ = $-\frac{8}{\sqrt{65}}$ , tanФ = -8, secФ = $\sqrt{65}$ , cscФ = $-\frac{\sqrt{65}}{8}$ , cotФ = $-\frac{1}{8}$

Step-by-step explanation:

If a point (x, y) lies on the terminal side of angle Ф in standard position, then the six trigonometry functions are:

cosФ = $\frac{x}{r}$
sinФ = $\frac{y}{r}$
tanФ = $\frac{y}{x}$
secФ = $\frac{r}{x}$
cscФ = $\frac{r}{y}$
cotФ = $\frac{x}{y}$

Where r = $\sqrt{x^{2}+y^{2} }$ (the length of the terminal side from the origin to point (x, y)

You should find the quadrant of (x, y) to adjust the sign of each function

∵ Point (1, -8) lies on the terminal side of angle Ф in standard position

∵ x is positive and y is negative

→ That means the point lies on the 4th quadrant

∴ Angle Ф is on the 4th quadrant

∵ In the 4th quadrant cosФ and secФ only have positive values

∴ sinФ, secФ, tanФ, and cotФ have negative values

→ let us find r

∵ r = $\sqrt{x^{2}+y^{2} }$

∵ x = 1 and y = -8

∴ r = $\sqrt{x} \sqrt{(1)^{2}+(-8)^{2}}=\sqrt{1+64}=\sqrt{65}$

→ Use the rules above to find the six trigonometric functions of Ф

∵ cosФ = $\frac{x}{r}$

∴ cosФ = $\frac{1}{\sqrt{65}}$

∵ sinФ = $\frac{y}{r}$

∴ sinФ = $-\frac{8}{\sqrt{65}}$

∵ tanФ = $\frac{y}{x}$

∴ tanФ = $-\frac{8}{1}$ = -8

∵ secФ = $\frac{r}{x}$

∴ secФ = $\frac{\sqrt{65}}{1}$ = $\sqrt{65}$

∵ cscФ = $\frac{r}{y}$

∴ cscФ = $-\frac{\sqrt{65}}{8}$

∵ cotФ = $\frac{x}{y}$

∴ cotФ = $-\frac{1}{8}$

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