Prove that Sin(-1) 1/√5 + sin (-1) 2/√5=π/2
Answers
Sin-14/5 + Sin-15/13 + Sin-116/65
Sin-1 [4/5 √(1-(5/13)2) + 5/13 √(1-(4/5)2)] + Sin-116/65 [∴Sin-1x + Sin-1y = Sin-1(x√(1-y2) + y√(1-x2)) ]
Sin-1[(4/5 * 12/13) + (5/13 * 3/5)] + Sin-116/65
Sin-1(63/65) + Sin-116/65
Sin-1 [63/65 √(1-(16/65)2) + 16/65 √(1-(63/65)2)]
Sin-1[(63/65 * 63/65) + (16/65 * 16/65)]
Sin-1[3969/4225 + 256/4225]
Sin-1(1)
π/2..............ans
Solution:
\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)+\sin ^{-1}\left(\frac{16}{65}\right)=\frac{\pi}{2}sin
−1
(
5
4
)+sin
−1
(
13
5
)+sin
−1
(
65
16
)=
2
π
\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{16}{65}\right)⇒sin
−1
(
5
4
)+sin
−1
(
13
5
)=
2
π
−sin
−1
(
65
16
)
\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\cos ^{-1}\left(\frac{16}{65}\right)⇒sin
−1
(
5
4
)+sin
−1
(
13
5
)=cos
−1
(
65
16
)
We just prove \bold{\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\cos ^{-1}\left(\frac{16}{65}\right)}sin
−1
(
5
4
)+sin
−1
(
13
5
)=cos
−1
(
65
16
)
We take left hand side =\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=sin
−1
(
5
4
)+sin
−1
(
13
5
)
Let \sin ^{-1}\left(\frac{4}{5}\right)=x \text { and } \sin ^{-1}\left(\frac{5}{13}\right)=ysin
−1
(
5
4
)=x and sin
−1
(
13
5
)=y
\Rightarrow \sin x=\frac{4}{5} \text { and } \sin y=\frac{5}{13}⇒sinx=
5
4
and siny=
13
5
Now, \cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}cosx=
1−sin
2
x
=
1−
25
16
=
5
3
Now, \cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-\frac{25}{169}}=\frac{12}{13}cosy=
1−sin
2
y
=
1−
169
25
=
13
12
Now, \bold{\cos (x+y)=\cos x \cdot \cos y-\sin x \cdot \sin y}cos(x+y)=cosx⋅cosy−sinx⋅siny
\Rightarrow \cos (x+y)=\frac{3}{5} \times \frac{12}{13}-\frac{4}{5} \times \frac{5}{13}⇒cos(x+y)=
5
3
×
13
12
−
5
4
×
13
5
\Rightarrow \cos (x+y)=\frac{36}{65}-\frac{20}{65}⇒cos(x+y)=
65
36
−
65
20
\Rightarrow \cos (x+y)=\frac{16}{65}⇒cos(x+y)=
65
16
\Rightarrow x+y=\cos ^{-1}\left(\frac{16}{65}\right)⇒x+y=cos
−1
(
65
16
)
\bold{\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\cos ^{-1}\left(\frac{16}{65}\right)}⇒sin
−1
(
5
4
)+sin
−1
(
13
5
)=cos
−1
(
65
16
) = Right hand side.
Hence proved.