Let sin(−θ)=−3/5 and tanθ>0. What is the value of cos(−θ)?

A 3/5

B−3/5

C 4/5

D −4/5

Question

contestada

Respuesta :

WaywardDelaney WaywardDelaney · Answer 1 · 2019-12-18T10:19:23+01:00

Answer:

The value of Cos (-Ф) [tex]\dfrac{\textrm 4}{\textrm 5}[/tex] .

Step-by-step explanation:

Given Trigonometric function as :

sin( - Ф ) = [tex]\frac{- 3}{5}[/tex]

- sin Ф = [tex]\frac{- 3}{5}[/tex]

So, sin Ф = [tex]\frac{ 3}{5}[/tex]

Now, as sin Ф = [tex]\dfrac{\textrm perpendicular}{\textrm hypotenuse}[/tex]

So , [tex]\dfrac{\textrm perpendicular}{\textrm hypotenuse}[/tex] = [tex]\frac{ 3}{5}[/tex]

So, perpendicular = 3

And hypotenuse = 5

Now, From Pythagoras Theorem

Base ² = Hypotenuse² - Perpendicular²

Or, Base ² = 5² - 3²

Or, Base ² = 25 - 9

Or, Base ² = 16

∴ Base = [tex]\sqrt{16}[/tex]

I.e Base = 4

Now, Cos Ф = [tex]\dfrac{\textrm base}{\textrm hypotenuse}[/tex]

So, Cos Ф = [tex]\dfrac{\textrm 4}{\textrm 5}[/tex]

Now , Since

Cos ( - Ф ) = Cos Ф

So, Cos ( - Ф ) = Cos Ф = [tex]\dfrac{\textrm 4}{\textrm 5}[/tex]

Hence The value of Cos (-Ф) [tex]\dfrac{\textrm 4}{\textrm 5}[/tex] . Answer