Question

if sin a + cos a =7/5 and sina cosa =12/25 find the value of sin a and cos a
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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

Answer:

sin a =4/5

cos a =3/5

Step-by-step explanation:

(sin a - cos a)² = sin²a -2sin a.cos a+cos²a

=1 - (2×12/25)

= (25-24/25)

=1/25

sin a - cos a =√(1/25)

±1/5

Now

sin a + cos a + sin a - cos a = (7/5)+(1/5)

2sin a = 8/5

sin a = 8/10 = 4/5

cos a =(7/5)-(8/10)

=14-8/10

=6/10= 3/5

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