The value of
sin(n+1) A-sin(n-1) A
cos(n+1) A+ cos(n-1) A
is
Answers
Answer:
Question:
Prove that:
s
i
n
(
n
+
1
)
A
s
i
n
(
n
−
1
)
A
+
c
o
s
(
n
+
1
)
A
c
o
s
(
n
−
1
)
A
=
c
o
s
2
A
.
Sum and Difference Formulas:
The sum and difference formulas in Trigonometry are:
sin
(
A
+
B
)
=
sin
A
cos
B
+
cos
A
sin
B
sin
(
A
−
B
)
=
sin
A
cos
B
−
cos
A
sin
B
cos
(
A
+
B
)
=
cos
A
cos
B
−
sin
A
sin
B
cos
(
A
−
B
)
=
cos
A
cos
B
+
sin
A
sin
B
In our problem, we will start with the left-hand side, apply the sum/difference formulas to arrive at the right-hand side.
We will start with the left-hand side.
$$\begin{align} \text{Left-hand side }&=\sin (n+1)A \sin (n-1)A + \cos (n+1)A \cos(n-1)A\\[0.4cm] & = \sin...
HERE IS YOUR ANSWER MATE.........
cos(A-B)=
cosA cosB +sinAsinB
Now
Let A = (n+1)A
B= (n-1)A
Now by 1st equation we can say that
cos [(n+1)A - (n-1)A]
=sin(n+1)A.sin(n-1)A+cos(n+1) A.cos(n-1)A
Left hand side
= cos [An+A-(An-A)]
cos =2A
HENCE PROVED..........