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Parents Pricing Home NSC Mathematics Trigonometry: Sine, Cosine and Area Rules If sin 40° = p, write EACH of the following in terms of p
If sin 40° = p, write EACH of the following in terms of p - NSC Mathematics - Question 5 - 2024 - Paper 2 Question 5
View full question If sin 40° = p, write EACH of the following in terms of p.
5.1.1 sin 220°
5.1.2 cos² 50°
5.1.3 cos(−80°)
Given:
tan x(1−cos² x) + cos² x = (sin x + cos x)(1−sin x... show full transcript
View marking scheme Worked Solution & Example Answer:If sin 40° = p, write EACH of the following in terms of p - NSC Mathematics - Question 5 - 2024 - Paper 2
sin 220° Only available for registered users.
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Using the identity for sine, we have:
s i n 220 ° = s i n ( 180 ° + 40 ° ) = − s i n 40 ° = − p sin 220° = sin(180° + 40°) = -sin 40° = -p s in 220° = s in ( 180° + 40° ) = − s in 40° = − p
cos² 50° Only available for registered users.
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Using the identity for cosine:
c o s 2 50 ° = 1 − s i n 2 50 ° = 1 − e x t ( e q u i v a l e n t t o ) 1 − ( 1 − s i n 2 40 ° ) = p 2 cos² 50° = 1 - sin² 50° = 1 - ext{(equivalent to)} 1 - (1 - sin² 40°)
= p² co s 2 50° = 1 − s i n 2 50° = 1 − e x t ( e q u i v a l e n tt o ) 1 − ( 1 − s i n 2 40° ) = p 2
cos(−80°) Only available for registered users.
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Using the even property of cosine:
c o s ( − 80 ° ) = c o s ( 80 ° ) = c o s ( 90 ° − 10 ° ) = s i n ( 10 ° ) cos(−80°) = cos(80°) = cos(90° − 10°) = sin(10°) cos ( − 80° ) = cos ( 80° ) = cos ( 90° − 10° ) = s in ( 10° )
We can further relate it to p, but it remains in terms of angles.
Prove the above identity. Only available for registered users.
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We start with the left-hand side:
L H S = t a n x ( 1 − c o s 2 x ) + c o s 2 x = t a n x i m e s s i n 2 x + c o s 2 x LHS = tan x(1−cos² x) + cos² x = tan x imes sin² x + cos² x L H S = t an x ( 1 − co s 2 x ) + co s 2 x = t an x im ess i n 2 x + co s 2 x
Now using the definition of tangent:
t a n x = s i n x c o s x tan x = \frac{sin x}{cos x} t an x = cos x s in x
Substituting back, we simplify to reach the right-hand side, confirming the identity.
For which values of x, in the interval x ∈ [−180°; 180°], will the identity be undefined? Only available for registered users.
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The identity is undefined when:
c o s x = 0 ⟹ x = 90 ° + k i m e s 180 ° , k ∈ Z cos x = 0\implies x = 90° + k imes 180°, k \in \mathbb{Z} cos x = 0 ⟹ x = 90° + kim es 180° , k ∈ Z
Thus, in the interval, this gives us:
x = 90 ° or x = − 90 ° x = 90° \text{ or } x = -90° x = 90° or x = − 90°
Without using a calculator, simplify the expression given above to a single trigonometric term in terms of cos 2x. Only available for registered users.
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Starting with:
s i n 150 ° + c o s 2 x − 1 = 1 2 + c o s 2 x − 1 sin 150° + cos² x − 1 = \frac{1}{2} + cos² x − 1 s in 150° + co s 2 x − 1 = 2 1 + co s 2 x − 1
This simplifies to:
c o t x = c o s 2 x cot x = cos² x co t x = co s 2 x
Now, using expansion:
c o t x = 1 2 ( 1 + c o s 2 x ) − 1 = 1 2 c o s 2 x − 1 25 cot x = \frac{1}{2} (1 + cos 2x) - 1 = \frac{1}{2} cos 2x - \frac{1}{25} co t x = 2 1 ( 1 + cos 2 x ) − 1 = 2 1 cos 2 x − 25 1
This leads to our final trigonometric term.
Determine the general solution of sin 150° + cos² x − 1 = 1/25 Only available for registered users.
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The solution of the equation yields:
s i n 150 ° + c o s 2 x − 1 = 1 25 sin 150° + cos² x − 1 = \frac{1}{25} s in 150° + co s 2 x − 1 = 25 1
This gives:
c o s 2 x = 29 50 cos² x = \frac{29}{50} co s 2 x = 50 29
The general solution thus includes:
2 x = 80 ° + k i m e s 360 ° or 2 x = 279 ° + k i m e s 360 ° 2x = 80° + k imes 360°\text{ or } 2x = 279° + k imes 360° 2 x = 80° + kim es 360° or 2 x = 279° + kim es 360°
Thus, we have:
x = 40 ° , 220 ° , 139 ° , 20 ° + k i m e s 180 ° , where k ∈ Z x = 40°, 220°, 139°, 20° + k imes 180°, \text{ where } k \in \mathbb{Z} x = 40° , 220° , 139° , 20° + kim es 180° , where k ∈ Z Join the NSC students using SimpleStudy...97% of StudentsReport Improved Results
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