Question

Will mark brainliest! Can something pls help me answer these showing there work? I offer 15 points

Answer 1

Answer:

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 positive
• At the second quadrant, only the sine trignometric ratio is positive
• At 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) = $-\frac{\sqrt{3} }{2}$