If cos theta+sin theta =root 2 90-theta show that sin theta -cos theta =root 2 cos theta
Answers
I think question is incomplete and it is like this
If Cosθ + Sinθ = √2 Cos ( 90° - θ ) then show that
Sinθ - Cosθ = √2 Cosθ
To prove --->
ATQ ,
Cosθ + Sinθ = √2 Cos(90° - θ )
We know that, Cos ( 90° -θ )= Cosθ , so
=> Cosθ + Sinθ = √2 Sinθ
=> Cosθ = √2 Sinθ - Sinθ
Taking Sinθ common in RHS
=> Cosθ = ( √2 - 1 ) Sinθ
Multiplying by (√2 + 1 ) in both sides
=> ( √2 + 1 ) Cosθ = (√2 +1) (√2 - 1) Sinθ
We have an identity as follows
(a + b ) (a - b ) = a² - b² , using it here
=> (√2 + 1 ) Cosθ = { ( √2 )² - ( 1 )² } Sinθ
=> (√2 + 1 ) Cosθ = ( 2 - 1 ) Sinθ
=> (√2 + 1 ) Cosθ = Sinθ
=> √2 Cosθ + Cosθ = Sinθ
=> √2 Cosθ = Sinθ - Cosθ
=> Sinθ - Cosθ = √2 Cosθ
Additional information--->
1) Sin²θ + Cos²θ = 1
2) 1 + tan²θ = Sec²θ
3) 1 + Cot²θ = Cosec²θ
Answer:
Step-by-step explanation:
If Cosθ + Sinθ = √2 Cos ( 90° - θ ) then show
that
Sinθ - Cosθ = √2 Cosθ
SOLUTION ;
We know that,
Cos ( 90° -θ )= Cosθ , so
=> Cosθ + Sinθ = √2 Sinθ
=> Cosθ = √2 Sinθ - Sinθ
Taking Sinθ common in RHS
=> Cosθ = ( √2 - 1 ) Sinθ
Multiplying Both sides by (√2 + 1 )
=> ( √2 + 1 ) Cosθ = (√2 +1) (√2 - 1) Sinθ
WE KNOW THAT
(a + b ) (a - b ) = a² - b² , using it here
=> (√2 + 1 ) Cosθ = { ( √2 )² - ( 1 )² } Sinθ
=> (√2 + 1 ) Cosθ = ( 2 - 1 ) Sinθ
=> (√2 + 1 ) Cosθ = Sinθ
=> √2 Cosθ + Cosθ = Sinθ
=> √2 Cosθ = Sinθ - Cosθ
=> Sinθ - Cosθ = √2 Cosθ