Mathematics, 24.01.2020 04:31, gracie2492

Will mark brainliest! can something pls me answer these showing there work? i offer 15 points

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answered: taijaababy

a) Third Quadrant

b) 7π/4 --> Option (4)

c) $-\frac{\sqrt{3} }{2}$ --> Option (1)

d) 1 --> Option (1)

e) $\frac{\sqrt{2} }{2}$ --> Option (2)

f) - $\frac{1}{2}$ --> Option (2)

g) $\frac{3}{2}$ --> Option (1)

h) $-\frac{\sqrt{3} }{2}$ --> Option(2)

Step-by-step explanation:

Ok, lets properly define some technical term here.

The terminal side of an angle is the side of the line after that it has made a turn (angle). I will drive my point home with the attachment to this solution

The initial side of an angle is the side of the line before the line made a turn(angle)

a) 1 complete revolution = $360^{0}$ = 2π rads

we can convert the radians to degrees using the above conversion rate

=> $\frac{7π}{6} rad \to degrees$ will be: $\frac{\frac{7π}{6} * 360}{2π}$

solving the expression above, 420π/2π = $210^{0}$

From the value of the angle in degree and having in mind that

$0^{0} - 90^{0} \to first \ quadrant\\ \\91^{0} - 180^{0} \to second\ quadrant\\\\181^{0} - 270^{0} \to third\ quadrant\\\\271^{0} - 360^{0} \to fourth\ quadrant$

$\frac{7π}{6} rad = 210^{0} \ is \ in \ third \ quadrant\\$

b) Co-terminal angles are angles which share the same initial and terminal side

To find the co-terminal of an angle we add or subtract 360 to the value if in degrees or 2π if in radians. From the value we want to find its co-terminal, because of the presence of π, its value is in radians and as such we add or subtract 2π from the value. If we perform subtraction, the negative co-terminal of the angle has been evaluated and the positive co-terminal is evaluated if we perform addition.

So, to get the positive co-terminal of -π/4, we add 2π and doing that, we get:

2π - π/4 = 7π/4

c) The value of sin(π/3) * cos(π) is ?

Applying special angle properties: (More on the special angle in the diagram attached to this solution)

sin(π/3) = $\frac{\sqrt{3} }{2}$

cos(π) = -1

substituting the values above into the expression, we have:

$\frac{\sqrt{3} }{2} * -1 = -\frac{\sqrt{3} }{2}$

d) if $f(x) = sin^{2}x + cos^{2} x$ , f(π/4) = ?

In trignometry, $sin^{2}x = (sin(x))^{2} ;\ cos^{2}x = (cos(x))^{2}$

Applying special angle properties again,

sin(π/4) = $\frac{\sqrt{2} }{2}$

cos(π/4) = $\frac{\sqrt{2} }{2}$

The expression becomes $(\frac{\sqrt{2} }{2} )^{2} + (\frac{\sqrt{2} }{2} )^{2}$ . Simplifying, we get:

2/4 + 2/4 = 1/2 + 1/2 = 1

e) cos(3π/4)

3π/4 is not an acute angle(angle < less than π/2 rad) and as such, we need to get its related acute angle. Now 3π/4 rads is in the second quadrant, this means that we will have to subtract 3π/4 from π to get the related acute angle.

π - 3π/4 = π/4

so instead of working with 3π/4, we work with its related acute angle which is π/4

cos(3π/4) is equivalent to cos(π/4) = $\frac{\sqrt{2} }{2}$ (special angle properties)

f) sin(11π/6)

11π/6 is not an acute angle(angle less than π/2 rad) and it is in the fourth quadrant. This means that to get its related acute angle, we have to subtract it from 2π

2π - 11π/6 = π/6

sin(11π/6) is equivalent to -sin(π/6) = -1/2 (special angle properties).

Note that there is a minus in the answer. That had nothing to do with the special angle properties but rather, the fact that:

At the fourth quadrant, only the cosine trignometric ratio is positive At the first quadrant, all trignometric ratios are positiveAt the second quadrant, only the sine trignometric ratio is positiveAt the third quadrant, only the tangent trignometric ratio is positive

g) sin(π/6) + tan(π/4)

using special angle properties:

sin(π/6) = 1/2 and tan(π/4) = 1

the expression simplifies to: 1/2+1 = 3/2

h) cos(4π/3)

4π/3 is not an acute angle and it is in the third quadrant

To get its related acute angle, we have to subtract it from 3π/2

3π/2 - 4π/3 = π/6

so, cos(4π/3) = -cos(π/6) (The negative value is because of the fact that at the third quadrant, only the tangent trignometric ratio is positive)

using special angle properties, -cos(π/6) = $-\frac{\sqrt{3} }{2}$