PLEASE SOLVE THIS EQUATION

MATH HELP PLZ!!!

RoseWind [281]

Answer:

a) $tan (157.5) = \frac{1-cos 315}{sin315}$

b)

$sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

c)

$sin^{2} (157.5) = \frac{1-cos (315) }{2}$

d)

cos 330° = 1- 2 sin² (165°)

Step-by-step explanation:

Step(i):-

By using trigonometry formulas

a)

cos2∝ = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ = 2 cos² ∝/2

$cos^{2} (\frac{\alpha }{2}) = \frac{1+cos\alpha }{2}$

b)

cos2∝ = 1- 2 sin² ∝

cos∝ = 1- 2 sin² ∝/2

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

Step(i):-

Given

$tan\alpha = \frac{sin\alpha }{cos\alpha }$

we know that trigonometry formulas

$sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

1- cos∝ = 2 sin² ∝/2

Given

$tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }$

put ∝ = 315

$tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }$

multiply with ' 2 sin (∝/2) both numerator and denominator

$tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2} }{2sin(\frac{315}{2} cos(\frac{315}{2}) }$

Apply formulas

$sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

1- cos∝ = 2 sin² ∝/2

now we get

$tan (157.5) = \frac{1-cos 315}{sin315}$

b)

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

put ∝ = 330° above formula

$sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}$

$sin^{2} (165) = \frac{1-cos (330) }{2}$

$sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

c )

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

put ∝ = 315° above formula

$sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}$

$sin^{2} (157.5) = \frac{1-cos (315) }{2}$

d)

cos∝ = 1- 2 sin² ∝/2

put ∝ = 330°

$cos 330 = 1 - 2sin^{2} (\frac{330}{2} )$

cos 330° = 1- 2 sin² (165°)

3 0

3 years ago

What is 3 = -2v − v v=

horsena [70]

Answer:

-1 =v

Step-by-step explanation:

3 = -2v − v

Combine like terms

3 = -3v

Divide each side by -3

3/-3 = -3v/-3

-1 =v

8 0

3 years ago

Read 2 more answers

Use the figure below for Exercises 1-8. Note that RN pierces the plane at N. It is not coplanar with V.

Serga [27]

Answer:

1. Two segments are MC and AX

2. Point N

3. Three collinear points are points A, point N and point X

4. Two other names for V are quadrilateral and rectangle

5. No N, M, and X are in a different plane from R

6. Two rays are ray NR and ray NA

7. The pair of opposite rays with end point N are ray NA and ray NX and ray NM and ray NC

8. Three lines are shown in the figure

Step-by-step explanation:

1) A segment is a line portion with two end points

2) A point marks a position

3) Collinear points are points on the same line

3) A quadrilateral is a four sided figure while rectangle has 90° interior angles

4) A ray is an line extending from a point without ending

5) Opposite rays are rays extending from the same point but in opposite directions

6) A line is a straight one dimensional geometric figure that has no width but infinite number of points extending without end in both directions

4 0

3 years ago

(9y® + 2)2 = combine like terms

Vsevolod [243]

Answer:

81y^2+36y+4

Step-by-step explanation:

First: 9y^2= 81y

Next: 2^2= 4

Then: 9y times 2= 18

Last: 18 times 2= 36y

(this : ^2 means squared in case you did not know)

4 0

4 years ago

My question is when multiplying scientific notations, after taking say, (8.2 * 4.5) and (10^9 * 10^-5) are you multiplying the 8

Kazeer [188]

Answer:you are adding them

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers