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Question 0511
Sigma Notation
2011 Paper 1 Question 6 Variant
Question
Generate new
(a)
Using the formulae for
sin
(
A
±
B
)
,
{\sin(A \pm B),}
sin
(
A
±
B
)
,
prove that
sin
(
r
+
1
2
)
θ
−
sin
(
r
−
1
2
)
=
2
cos
r
θ
sin
1
2
θ
.
\sin {\textstyle (r + \frac{1}{2})} \theta - \sin {\textstyle (r - \frac{1}{2})} = 2 \cos r \theta \sin {\textstyle \frac{1}{2} \theta}.
sin
(
r
+
2
1
)
θ
−
sin
(
r
−
2
1
)
=
2
cos
r
θ
sin
2
1
θ
.
[2]
(b)
Hence find a formula for
∑
r
=
1
n
cos
r
θ
{\displaystyle \sum_{r=1}^n \cos r \theta}
r
=
1
∑
n
cos
r
θ
in terms of
sin
(
n
+
1
2
)
θ
{\sin (n+\frac{1}{2}) \theta}
sin
(
n
+
2
1
)
θ
and
sin
1
2
θ
.
{\sin \frac{1}{2}\theta.}
sin
2
1
θ
.
[3]
Answer
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