Question

If A lies between 270° and 370° and sin A = -7/25 , then

1) sin2A = -336/625
2) cos A/2 = √(2)/5
3) tan A/2 = -1/7
4) sinA/2 = -√(2)/10





PLEASE SOLVE THE ABOVE QUESTION FASTLY AND CORRECTLY



3RD QUESTION IN THE ATTACHED IMAGE BELOW

Answer 1

Answer:

$sin2A=\frac{-336}{625}$

option (1), (3) are correct

Step-by-step explanation:

Formula used:

$sin^{A}+cos^2{A}=1$

$sin2A=2\:sinA\:cosA$

Given:

$sinA=\frac{-7}{25}$

since A lies between 270° and 360°, A lies in fourth quadrant

so, cosA is positive

Now,

$sin^2A+cos^2A=1$

$(\frac{-7}{25})^2+cos^2A=1$

$\frac{49}{625}+cos^2A=1$

$cos^2A=1-\frac{49}{625}$

$cos^2A=\frac{625-49}{625}$

$cos^2A=\frac{576}{625}$

$cosA=\sqrt{\frac{576}{625}}$

$\implies\:$$cosA=\frac{24}{25}$

$sin2A=2\:sinA\:cosA$

$sin2A=2(\frac{-7}{25})(\frac{24}{25})$

$sin2A=\frac{-336}{625}$

consider

$sinA=\frac{2\:tan\frac{A}{2}}{1+tan^2\frac{A}{2}}$

$\frac{-7}{25}=\frac{2\:tan\frac{A}{2}}{1+tan^2\frac{A}{2}}$

$-7-7tan^2\frac{A}{2}=50\:tan\frac{A}{2}$

Rearranging terms we get

$7tan^2\frac{A}{2}+50\:tan\frac{A}{2}+7=0$

$(tan\frac{A}{2}+7)(7tan\frac{A}{2}+1)=0$

$tan\frac{A}{2}=-7,\frac{-1}{7}$

since A lies in IV quadrant, $\frac{A}{2}$ lies in II quadrant

$\implies\:tan\frac{A}{2}=\frac{-1}{7}$

Answer 2

Answer:

sin2A = -336/625

Tan(A/2)  = -1/7

Step-by-step explanation:

If A lies between 270° and 370° and sin A = -7/25 , then

1) sin2A = -336/625

2) cos A/2 = √(2)/5

3) tan A/2 = -1/7

4) sinA/2 = -√(2)/10

A lies between  270° and 360°

=> A/2 lies between  135° & 180° ( Quadrant II)

=> Cos(A/2) would be -ve  hence Cos(A/2) ≠ √(2)/5

=> Sin(A/2) would be +ve  hence Sin(A/2) ≠ -√(2)/10

tan A/2 would be -ve  so value need to be checked

2A = 540° to 720° => 180° to 360°

=> Sin2A would be - ve  so value need to be checked

Sin2A = 2SinACosA

SinA = -7/25

CosA = √(1 - Sin²A) = √(1 - (-7/25)²)  =  √(625 - 49)/625 = √576/625 = ±24/25

as A is in 4th Quadrent so cosA is +ve

=> CosA = 24/25

Sin2A = 2(-7/25)(24/25) = - 336/625  

hence sin2A = -336/625

TanA = SinA/CosA = (-7/25)/(24/25) = -7/24

TanA = 2Tan(A/2)(1 - Tan²(A/2))

=> -7/24 = 2Tan(A/2)(1 - Tan²(A/2))

=> 7Tan²(A/2) - 7 = 48Tan(A/2)

=>  7Tan²(A/2) - 48Tan(A/2) - 7 = 0

=> 7Tan²(A/2) - 49Tan(A/2) + Tan(A/2)- 7 = 0

=> 7Tan(A/2)(Tan(A/2) - 7) + 1(Tan(A/2) -7) = 0

=> (7Tan(A/2) + 1)(Tan(A/2) - 7) = 0

=> Tan(A/2)  = -1/7  or Tan(A/2) = 7

Tan(A/2) should be -ve as lies in 2nd Quadrant

=> Tan(A/2)  = -1/7

option 1 & 3 are correct

sin2A = -336/625

Tan(A/2)  = -1/7