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secondary 4 | A Maths
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@Eric Nicholas K : I tried using R formulae but could not get the correct ans … could u help me to spot what went wrong?
Sorry I accidentally deleted the previous post about this question...
Date Posted: 3 years ago
Views: 368
Firstly, 2 cos²x - 1 = cos 2x and not sin 2x
Secondly, when you apply the R formula, your argument/angle ( θ ) = 2x, not x.
3 sin 2x + cos 2x = 1
R = √(3² + 1²) = √10
Basic angle α = tan-¹ (⅓) ≈ 18.4349°
So, rewrite as :
√10 sin (2x + 18.4349°) ≈ 1
sin (2x + 18.4339°) ≈ 1/√10
2x + 18.4349° ≈ sin-¹ (1/√10)
Basic angle θ = sin-¹ (1/√10) ≈ 18.4349°
Since the sine is positive, look at the 1st and 2nd quadrant.
But firstly, the range needs to be changed.
0° < x < 360° so 0° < 2x < 720°
2x + 18.4349° ≈ 180° - 18.4349°, 360° + 18.4349°, 180° - 18.4349° + 360°
2x ≈ 143.1302°, 360°, 503.1302°
x ≈ 71.5651°, 180°, 251.5651°
x = 71.6°, 180° , 251.6° (1d.p)
3 sin 2x + cos 2x = 1
R = √(3² + 1²) = √10
Basic angle α = tan-¹ (⅓) ≈ 18.4349°
So, rewrite as :
√10 sin (2x + 18.4349°) ≈ 1
sin (2x + 18.4339°) ≈ 1/√10
2x + 18.4349° ≈ sin-¹ (1/√10)
Basic angle θ = sin-¹ (1/√10) ≈ 18.4349°
Since the sine is positive, look at the 1st and 2nd quadrant.
But firstly, the range needs to be changed.
0° < x < 360° so 0° < 2x < 720°
2x + 18.4349° ≈ 180° - 18.4349°, 360° + 18.4349°, 180° - 18.4349° + 360°
2x ≈ 143.1302°, 360°, 503.1302°
x ≈ 71.5651°, 180°, 251.5651°
x = 71.6°, 180° , 251.6° (1d.p)
But actually, why the R-formula? It's not really needed. The usual way is easier.
3 sin 2x + 2 cos² x = 2
3 (2 sin x cos x) + 2 cos² x = 2 sin² x + 2 cos² x
(Recall the double angle formula for sine)
Then, cancel out 2 cos² x on both sides and divide both sides by 2,
3 sin x cos x = sin² x
sin² x - 3 sin x cos x = 0
(sin x)(sin x - 3 cos x) = 0
sin x = 0
OR
sin x = 3 cos x
sin x / cos x = 3
tan x = 3
For sin x = 0 ,
x = 0°, 180°, 360°, 540°, 720°,...
But since 0° < x < 360°,
x = 180°
For tan x = 3,
Basic angle θ = tan-¹ (3) ≈ 71.565°
The tangent is positive so look at the 1st and 3rd quadrant.
x ≈ 71.565°, 180° + 71.565°
x ≈ 71.565°, 251.565°
x = 71.6°, 251.6° (1 d.p)
3 sin 2x + 2 cos² x = 2
3 (2 sin x cos x) + 2 cos² x = 2 sin² x + 2 cos² x
(Recall the double angle formula for sine)
Then, cancel out 2 cos² x on both sides and divide both sides by 2,
3 sin x cos x = sin² x
sin² x - 3 sin x cos x = 0
(sin x)(sin x - 3 cos x) = 0
sin x = 0
OR
sin x = 3 cos x
sin x / cos x = 3
tan x = 3
For sin x = 0 ,
x = 0°, 180°, 360°, 540°, 720°,...
But since 0° < x < 360°,
x = 180°
For tan x = 3,
Basic angle θ = tan-¹ (3) ≈ 71.565°
The tangent is positive so look at the 1st and 3rd quadrant.
x ≈ 71.565°, 180° + 71.565°
x ≈ 71.565°, 251.565°
x = 71.6°, 251.6° (1 d.p)
Ohhh I got it! Tysm!!
But I think I have difficulty applying the R formula :( why do we need to calculate the basic angle of "a" first? (The one with result 18.435°)
You mean, your teacher has not explained the R formula and how it works?
https://www.exampaper.com.sg/tuition-notes/a-maths-tips/trigonometry-type-r
https://brilliant.org/wiki/trigonometric-r-method/
You can check these out.
https://brilliant.org/wiki/trigonometric-r-method/
You can check these out.
thank you!!!
I'm sure my teacher did :" but I think I didn't understand the concept correctly :( sorry about that
I'm sure my teacher did :" but I think I didn't understand the concept correctly :( sorry about that
I'll go and read it up~
I posted a picture. Check the comments there
See 2 Answers
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Visualisation. Explanation in comments.
Date Posted:
3 years ago
Triangle BON, DOM, ADB, ACB and CDB are similar triangles.
∠ACB = θ
For triangle BON ,
sin θ = opposite/hypotenuse = OB/BN = BN / 6
BN = 6 sin θ
For triangle ACB,
cos θ = adjacent / hypotenuse = AC / AB = AC / 6
AC = 2 cos θ
AM
= CM + AC
= BN + AC (since CBMN is a rectangle, CM = BN)
= 6 sin θ + 2 cos θ
But now, look at triangle AOM.
sin ∠AOM = opposite/hypotenuse = AM / OA
sin (θ + α) = (6 sin θ + 2 cos θ) / OA
OA sin (θ + α) = 6 sin θ + 2 cos θ
How to find OA? Look at triangle AOB.
OA² = OB² + AB² (Pythagoras's Theorem)
OA² = 6² + 2²
OA = √(6² + 2²) = √40
How to find α ?
tan α = opposite / adjacent = AB/OB = 2/6 = ⅓
α = tan-¹ (⅓)
(The basic angle)
So, putting these together,
√(6² + 2²) sin (θ + tan-¹(⅓) ) = 6 sin θ + 2 cos θ
√40 sin (θ + tan-¹(⅓)) = 6 sin θ + 2 cos θ
∠ACB = θ
For triangle BON ,
sin θ = opposite/hypotenuse = OB/BN = BN / 6
BN = 6 sin θ
For triangle ACB,
cos θ = adjacent / hypotenuse = AC / AB = AC / 6
AC = 2 cos θ
AM
= CM + AC
= BN + AC (since CBMN is a rectangle, CM = BN)
= 6 sin θ + 2 cos θ
But now, look at triangle AOM.
sin ∠AOM = opposite/hypotenuse = AM / OA
sin (θ + α) = (6 sin θ + 2 cos θ) / OA
OA sin (θ + α) = 6 sin θ + 2 cos θ
How to find OA? Look at triangle AOB.
OA² = OB² + AB² (Pythagoras's Theorem)
OA² = 6² + 2²
OA = √(6² + 2²) = √40
How to find α ?
tan α = opposite / adjacent = AB/OB = 2/6 = ⅓
α = tan-¹ (⅓)
(The basic angle)
So, putting these together,
√(6² + 2²) sin (θ + tan-¹(⅓) ) = 6 sin θ + 2 cos θ
√40 sin (θ + tan-¹(⅓)) = 6 sin θ + 2 cos θ
Sorry, ∠CAB = θ, not ∠ACB
Thx for clearing my doubts!! I got it now
No prob
Hi Mike, see your question's comments section for working and explanation.
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