J's answer to Rella's Secondary 4 A Maths Singapore question.
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p = 7/12
Usually, you have to use R-formula to convert sin x + cos x into a single trigonometric term first. Then solve for it.
But I will show a different method.
sin x + cos x = 1/√2
Square both sides,
(sin x + cos x)² = (1/√2)²
sin² x + 2 sin x cos x + cos² x = ½
1 + 2 sin x cos x = ½
2 sin x cos x = -½
sin 2x = -½ → negative (3rd or 4th quadrant)
Basic angle θ = sin-¹ (½) = π/6 rad
2x = -π/6 rad, (π/6 + π) rad, (2π - π/6) rad,...
2x = -π/6 rad, 7π/6 rad, 11π/6 rad
x = -π/12 rad, 7π/12 rad, 11π/12 rad
Smallest positive x = 7π/12 rad = 7/12 π rad
So smallest p = 7/12
Usually, you have to use R-formula to convert sin x + cos x into a single trigonometric term first. Then solve for it.
But I will show a different method.
sin x + cos x = 1/√2
Square both sides,
(sin x + cos x)² = (1/√2)²
sin² x + 2 sin x cos x + cos² x = ½
1 + 2 sin x cos x = ½
2 sin x cos x = -½
sin 2x = -½ → negative (3rd or 4th quadrant)
Basic angle θ = sin-¹ (½) = π/6 rad
2x = -π/6 rad, (π/6 + π) rad, (2π - π/6) rad,...
2x = -π/6 rad, 7π/6 rad, 11π/6 rad
x = -π/12 rad, 7π/12 rad, 11π/12 rad
Smallest positive x = 7π/12 rad = 7/12 π rad
So smallest p = 7/12
Date Posted:
3 years ago
Edit : note that some values are not acceptable (eg x = 11π/12, x = 19π/12, etc) as they would give sin x + cos x = -1/√2 .
This is because we squared both sides earlier on, so solutions for both the positive square root (1/√2) and negative square root (-1/√2) would appear.
You will need to check and reject certain values.
For R-formula method, this won't be necessary.
This is because we squared both sides earlier on, so solutions for both the positive square root (1/√2) and negative square root (-1/√2) would appear.
You will need to check and reject certain values.
For R-formula method, this won't be necessary.
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