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Parents Pricing Home SSCE HSC Mathematics Extension 2 Recursive formula proofs Question 6 (15 marks) Use a SEPARATE writing booklet
Question 6 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 6 - 2003 - Paper 1 Question 6
View full question Question 6 (15 marks) Use a SEPARATE writing booklet.
(a) (i) Prove the identity $ ext{cos}(a + b) + ext{cos}(a - b) = 2 ext{cos} a ext{cos} b$.
(ii) Hence fi... show full transcript
View marking scheme Worked Solution & Example Answer:Question 6 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 6 - 2003 - Paper 1
Prove the identity $ ext{cos}(a + b) + ext{cos}(a - b) = 2 ext{cos} a ext{cos} b$ Only available for registered users.
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To prove the identity, we start with the left-hand side:
e x t c o s ( a + b ) + e x t c o s ( a − b ) ext{cos}(a + b) + ext{cos}(a - b) e x t cos ( a + b ) + e x t cos ( a − b )
Using the sum-to-product identities, we can write:
ext{cos}(a + b) = rac{e^{i(a+b)} + e^{-i(a+b)}}{2}
ext{cos}(a - b) = rac{e^{i(a-b)} + e^{-i(a-b)}}{2}
Combining these gives:
= rac{e^{ia} (e^{ib} + e^{-ib}) + e^{-ia} (e^{-ib} + e^{ib})}{2} = ext{cos} a ext{cos} b.
Hence find $\int \text{cos} 3x \text{cos} 2x \, dx$ Only available for registered users.
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Using the identity from part (i):
cos 3 x cos 2 x = 1 2 ( cos ( 3 x + 2 x ) + cos ( 3 x − 2 x ) ) = 1 2 ( cos ( 5 x ) + cos ( x ) ) . \text{cos} 3x \text{cos} 2x = \frac{1}{2} \left( \text{cos}(3x + 2x) + \text{cos}(3x - 2x) \right) = \frac{1}{2} \left( \text{cos}(5x) + \text{cos}(x) \right). cos 3 x cos 2 x = 2 1 ( cos ( 3 x + 2 x ) + cos ( 3 x − 2 x ) ) = 2 1 ( cos ( 5 x ) + cos ( x ) ) .
Thus,
∫ cos 3 x cos 2 x d x = 1 2 ( ∫ cos ( 5 x ) d x + ∫ cos ( x ) d x ) . \int \text{cos} 3x \text{cos} 2x \, dx = \frac{1}{2} \left( \int \text{cos}(5x) \, dx + \int \text{cos}(x) \, dx \right). ∫ cos 3 x cos 2 x d x = 2 1 ( ∫ cos ( 5 x ) d x + ∫ cos ( x ) d x ) .
Integrating gives:
= 1 2 ( 1 5 sin ( 5 x ) + sin ( x ) + C ) = 1 10 sin ( 5 x ) + 1 2 sin ( x ) + C . = \frac{1}{2} \left( \frac{1}{5} \text{sin}(5x) + \text{sin}(x) + C \right) = \frac{1}{10} \text{sin}(5x) + \frac{1}{2} \text{sin}(x) + C. = 2 1 ( 5 1 sin ( 5 x ) + sin ( x ) + C ) = 10 1 sin ( 5 x ) + 2 1 sin ( x ) + C .
Find $s_3$ and $s_4$ Only available for registered users.
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Given s 1 = 1 s_1 = 1 s 1 = 1 s 2 = 2 s_2 = 2 s 2 = 2
s 3 = s 2 + ( 3 − 1 ) s 1 = 2 + 2 ⋅ 1 = 4. s_3 = s_2 + (3 - 1)s_1 = 2 + 2 \cdot 1 = 4. s 3 = s 2 + ( 3 − 1 ) s 1 = 2 + 2 ⋅ 1 = 4.
s 4 = s 3 + ( 4 − 1 ) s 2 = 4 + 3 ⋅ 2 = 10. s_4 = s_3 + (4 - 1)s_2 = 4 + 3 \cdot 2 = 10. s 4 = s 3 + ( 4 − 1 ) s 2 = 4 + 3 ⋅ 2 = 10.
Prove that $\sqrt{x + x^2} = \sqrt{x(x + 1)}$ for all real numbers $x \geq 0$ Only available for registered users.
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Starting with the left side:
Since $x \geq 0$, we can safely say:
$$\sqrt{x(1 + x)} = \sqrt{x(x + 1)}.$$
Thus the equality is confirmed.
Prove by induction that $s_n \geq \sqrt{n}$ for all integers $n \geq 1$ Only available for registered users.
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Base case: For n = 1 n=1 n = 1 s 1 = 1 ≥ 1 s_1 = 1 \geq \sqrt{1} s 1 = 1 ≥ 1
Inductive step: Assume true for n = k n=k n = k s k ≥ k s_k \geq \sqrt{k} s k ≥ k n = k + 1 n=k+1 n = k + 1
\geq \sqrt{k} + (k)\sqrt{k-1} \
\geq \sqrt{k + 1}.$$
Thus the statement holds for $n=k+1$.
Prove that $\frac{x+y}{2} \geq \sqrt{xy}$ Only available for registered users.
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This is proven by applying the AM-GM inequality:
x + y 2 ≥ x y \frac{x+y}{2} \geq \sqrt{xy} 2 x + y ≥ x y x + y ≥ 2 x y x + y \geq 2\sqrt{xy} x + y ≥ 2 x y
Rearranging gives:
( x − y ) 2 ≥ 0 , (x - y)^2 \geq 0, ( x − y ) 2 ≥ 0 ,
which is always true.
Prove that $a^2 + b^2 + c^2 \geq ab + ac + bc$ Only available for registered users.
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Using the three-variable version of Cauchy-Schwarz inequality:
( a 2 + b 2 + c 2 ) ( 1 + 1 + 1 ) ≥ ( a + b + c ) 2 , (a^2 + b^2 + c^2)(1+1+1) \geq (a + b + c)^2, ( a 2 + b 2 + c 2 ) ( 1 + 1 + 1 ) ≥ ( a + b + c ) 2 ,
This proves:
a 2 + b 2 + c 2 ≥ a b + a c + b c . a^2 + b^2 + c^2 \geq ab + ac + bc. a 2 + b 2 + c 2 ≥ ab + a c + b c .
Show that $a^2b^2 + c^2 \geq 2ab(a+b)$ Only available for registered users.
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By applying AM-GM on a 2 b 2 a^2b^2 a 2 b 2 c 2 c^2 c 2
a 2 b 2 + c 2 2 ≥ a 2 b 2 ⋅ c 2 = a b c 2 = 2 a b ( a + b ) . \frac{a^2b^2 + c^2}{2} \geq \sqrt{a^2b^2 \cdot c^2} = ab\sqrt{c^2} = 2ab(a + b). 2 a 2 b 2 + c 2 ≥ a 2 b 2 ⋅ c 2 = ab c 2 = 2 ab ( a + b ) .
Deduce that if $a + b + c = d$, then $a^2 + b^2 + c^2 \geq ab + ac + bc$ Only available for registered users.
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Replacing c c c d − a − b d - a - b d − a − b
a 2 + b 2 + ( d − a − b ) 2 ≥ a b + a c + b c , a^2 + b^2 + (d - a - b)^2 \geq ab + ac + bc, a 2 + b 2 + ( d − a − b ) 2 ≥ ab + a c + b c ,
Confirms that:
a 2 + b 2 + c 2 ≥ a b + a c + b c . a^2 + b^2 + c^2 \geq ab + ac + bc. a 2 + b 2 + c 2 ≥ ab + a c + b c .
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