Question

if sin theta + cos theta equals to 7 by 5 and sin theta into cos theta equals to 12 by 25 then find the values of sin theta and Cos theta​

Answer 1

Answer:

sinA = 3/5 or 4/5

cosA = 4/5 or 3/5

Step-by-step explanation:

  Let sinA be a and cosA.

Given,

 sinA + cosA = 7 / 5

          ⇒ a + b = 7 / 5

 sinA.cosA = 12 / 25

          ⇒ ab = ( 12 / 25 )

Square on both sides of a + b:

⇒ ( a + b )^2 = ( 7 / 5 )^2

⇒ a^2 + b^2 + 2ab = ( 7 / 5 )^2

⇒ a^2 + b^2 + 2(12/25) = (7/5)^2   { from above,ab=12/25}

⇒ a^2 + b^2 = (49/25) - 2(12/25)

⇒ a^2 + b^2 = (49-24)/25

⇒ a^2 + b^2 = (25/25)

   Adding - 2ab both sides :

⇒ a^2 + b^2 - 2ab = (25/35) - 2(12/25)

⇒ ( a - b )^2 = (25/25) - (24/25)

⇒ a - b = 1 / 5 of - 1 / 5

    now, a + b = 7 / 5

             a - b = 1 / 5

            2a = 8 / 5

              a = 4 / 5  

         Thus,

            a - 1 / 5 = b    ⇒ 4/5 - 1/5 = b    ⇒ 3/5 = b

Or   if a - b = - 1 / 5

          a + b = 7 / 5

          2a = 6/5

          a = 3/5

          So, b = 4/5

Hence,

 a = sinA = 3/5 or 4/5

 b = cosA = 4/5 or 3/5

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

if sin θ + cos θ equals to 7/5 and sin θ.cos θ equals to 12/25 then find the values of sin θ and Cos θ .

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• sin θ + cos θ = 7/5 ...........(1)
• sin θ .cos θ = 12/25 ..........(2)

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• Value of sin θ and cos θ

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

We know,

★(sin θ - cos θ) = √[(sin θ + cos θ)²-4sin θ .cos θ]

Keep value by equ(1) and equ(2),

➩(sin θ - cos θ) = √[(7/5)²-4×(12/25)]

➩(sin θ - cos θ) = √[49/25 - 48/25]

➩(sin θ - cos θ) = √[(49-48)/25]

➩(sin θ - cos θ) = √(1/25)

➩(sin θ - cos θ) = 1/5 ...............(3)

Add equ(1) and equ(3),

➩ 2.sin θ = 7/5 + 1/5

➩ 2.sin θ = (7+1)/5

➩ 2.sin θ = 8/5

➩ sin θ = 8/(5×2)

➩ sin θ = 4/5

keep this value equ(3),

➩ (4/5) - cos θ = 1/5

➩ cos θ = 4/5 - 1/5

➩ cos θ = (4-1)/5

➩ cos θ = 3/5

$\Large{\underline{\mathfrak{\bf{Thus}}}}$

• Value of sin θ = 4/5
• Value of cos θ = 3/5

$\Large{\underline{\underline{\mathfrak{\bf{Verification}}}}}$

Keep value of sin θ and cos θ in equ(1),

➩ (4/5) + (3/5) = 7/5

➩ (4+3)/5 = 7/5

➩ 7/5 = 7/5

L.H.S. = R.H.S.

Again,

Keep value of sin θ and cos θ in equ(2),

➩ (4/5).(3/5) = 12/25

➩ 12/25 = 12/25

LH.S=R.H.S

That's proved