What is :-
--> 1 + sinx and
--> 1 - sinx
Formula
Answers
Answered by
0
Let us generalise both the formula as 1±
1
±
s
i
n
x
.
=2(2)(2)
s
i
n
x
=
2
s
i
n
(
x
2
)
c
o
s
(
x
2
)
Also, we know, 2(2)+2(2)=1
s
i
n
2
(
x
2
)
+
c
o
s
2
(
x
2
)
=
1
∴1±=2(2)+2(2)±2(2)(2)
∴
1
±
s
i
n
x
=
s
i
n
2
(
x
2
)
+
c
o
s
2
(
x
2
)
±
2
s
i
n
(
x
2
)
c
o
s
(
x
2
)
We know, (±)2=2±2+2
(
a
±
b
)
2
=
a
2
±
2
a
b
+
b
2
Hence, 1±=((2)±(2))2
1
±
s
i
n
x
=
(
s
i
n
(
x
2
)
±
c
o
s
(
x
2
)
)
2
.
Hence, 1+=((2)+(2))2
1
+
s
i
n
x
=
(
s
i
n
(
x
2
)
+
c
o
s
(
x
2
)
)
2
and 1−=((2)−(2))2
1
−
s
i
n
x
=
(
s
i
n
(
x
2
)
−
c
o
s
(
x
2
)
)
2
Hope this helps! You can also get such questions solved and get the detailed solution within seconds using the Scholr app by just uploading a picture of the question and also get to be a part of an ever-growing community of students.
1
±
s
i
n
x
.
=2(2)(2)
s
i
n
x
=
2
s
i
n
(
x
2
)
c
o
s
(
x
2
)
Also, we know, 2(2)+2(2)=1
s
i
n
2
(
x
2
)
+
c
o
s
2
(
x
2
)
=
1
∴1±=2(2)+2(2)±2(2)(2)
∴
1
±
s
i
n
x
=
s
i
n
2
(
x
2
)
+
c
o
s
2
(
x
2
)
±
2
s
i
n
(
x
2
)
c
o
s
(
x
2
)
We know, (±)2=2±2+2
(
a
±
b
)
2
=
a
2
±
2
a
b
+
b
2
Hence, 1±=((2)±(2))2
1
±
s
i
n
x
=
(
s
i
n
(
x
2
)
±
c
o
s
(
x
2
)
)
2
.
Hence, 1+=((2)+(2))2
1
+
s
i
n
x
=
(
s
i
n
(
x
2
)
+
c
o
s
(
x
2
)
)
2
and 1−=((2)−(2))2
1
−
s
i
n
x
=
(
s
i
n
(
x
2
)
−
c
o
s
(
x
2
)
)
2
Hope this helps! You can also get such questions solved and get the detailed solution within seconds using the Scholr app by just uploading a picture of the question and also get to be a part of an ever-growing community of students.
Answered by
1
===> ( 1 + SinX ) ( 1 - SinX )
===> ( 1 ) 2 - Sin2X
【 ( a + b ) ( a - b ) = a2 - b2 】
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