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Sin x ( sin x + cos x ) = p

Sin² x + Sin x cos x = p

(1 - Cos 2 x ) / 2 + 1/2 Sin 2 x = p

1 - Cos 2 x + Sin 2 x = 2 p

Sin 2x - Cos 2 x = 2p - 1

Sin 2x - SIn (π/2 - 2x) = 2p - 1

the rule Sin A - Sin B = 2 Sin (A - B)/2 Cos (A+B)/2

So 2 Sin ( 2x - π/4) Cos π/4 = 2p -1

*Sin (2 x - π/4) = (2p -1) / √2*

Sin of an angle is always between + 1 and -1.

So* -1 ≤ (2p -1)/√2 ≤ 1*

2p -1 >= - √2 =>* p >= (1-√2)/2 *

and 2 p - 1 <= √2 =>** p <= (1+√2)/2**

*The limits for p , or the interval for p is : 0 ≤ p ≤ 1*

=========================================

Sin x ( sin x + cos x ) = p

Sin x (sin x + Sin (90 -x) ) = p

we use the rule Sin A + Sin B = 2 Sin (A + B)/2 Cos (A-B)/2

Sin x [ 2 Sin 45⁰ Cos (x - 45⁰) ]= p

2 Sin x Cos (x - 45°) = √2 p

Sin (2 x - 45°) + Sin 45° = √2 p

Sin (2x - 45°) = √2 p - 1/√2 = (2p -1)/√2

As Sine of an angle is always between -1 and 1, we get that,

2p -1 >= - √2 =>* p >= (1-√2)/2 *

and 2 p - 1 <= √2 =>* p <= (1+√2)/2*

====================================

if you know differentiation then:

Sin x ( sin x + cos x ) = p

Sin² x + sin x Cos x = p

Sin² x + 1/2 Sin 2x = p

dp / dx = deriviate wrt x = 2 Sin x + Cos 2 x

make the derivative = 0 to find maximum or minimum.

Hence, 2 Sin x + 1 - 2 Sin² x = 0

Sin² x - Sin x - 1/2 = 0

this is a quadratic equation : Sin x = [ 1 + - √(1+2) ] / 2

Sin x = (1 + √3 )/2 or (1 - √3)/2

= only (1 - √3)/2 as the other one is more than 1. so x is not real.

= -0.366

Then cos² x = 1 - sin² x = 1 - 1/4 [ 1 + 3 - 2√3 ]

= √3/2

Cos x = √(√3/2) = 0.9306

substitute the values in the expression to get the max or min value.

Sin² x + Sin x cos x = p

(1 - Cos 2 x ) / 2 + 1/2 Sin 2 x = p

1 - Cos 2 x + Sin 2 x = 2 p

Sin 2x - Cos 2 x = 2p - 1

Sin 2x - SIn (π/2 - 2x) = 2p - 1

the rule Sin A - Sin B = 2 Sin (A - B)/2 Cos (A+B)/2

So 2 Sin ( 2x - π/4) Cos π/4 = 2p -1

Sin of an angle is always between + 1 and -1.

So

2p -1 >= - √2 =>

and 2 p - 1 <= √2 =>

=========================================

Sin x ( sin x + cos x ) = p

Sin x (sin x + Sin (90 -x) ) = p

we use the rule Sin A + Sin B = 2 Sin (A + B)/2 Cos (A-B)/2

Sin x [ 2 Sin 45⁰ Cos (x - 45⁰) ]= p

2 Sin x Cos (x - 45°) = √2 p

Sin (2 x - 45°) + Sin 45° = √2 p

Sin (2x - 45°) = √2 p - 1/√2 = (2p -1)/√2

As Sine of an angle is always between -1 and 1, we get that,

2p -1 >= - √2 =>

and 2 p - 1 <= √2 =>

====================================

if you know differentiation then:

Sin x ( sin x + cos x ) = p

Sin² x + sin x Cos x = p

Sin² x + 1/2 Sin 2x = p

dp / dx = deriviate wrt x = 2 Sin x + Cos 2 x

make the derivative = 0 to find maximum or minimum.

Hence, 2 Sin x + 1 - 2 Sin² x = 0

Sin² x - Sin x - 1/2 = 0

this is a quadratic equation : Sin x = [ 1 + - √(1+2) ] / 2

Sin x = (1 + √3 )/2 or (1 - √3)/2

= only (1 - √3)/2 as the other one is more than 1. so x is not real.

= -0.366

Then cos² x = 1 - sin² x = 1 - 1/4 [ 1 + 3 - 2√3 ]

= √3/2

Cos x = √(√3/2) = 0.9306

substitute the values in the expression to get the max or min value.