If x and y are acute angles such that cosx + cosy =
and sin x + sin yr , then sin(x + y) equals to
Answers
Answer:
Given,
sin x =\frac{1}{\sqrt{5}}\text{ and }sin y = \frac{1}{\sqrt{10}}sinx=
5
1
and siny=
10
1
cos x = \sqrt{1-(\frac{1}{\sqrt{5}})^2}\text{ and }cos y = \sqrt{1-(\frac{1}{\sqrt{10}})^2}cosx=
1−(
5
1
)
2
and cosy=
1−(
10
1
)
2
( Because sin²A + cos²A = 1 ⇒ cos²A = 1 - sin²A ⇒ cos A = √(1-sin²A)
cos x = \sqrt{1-\frac{1}{5}}\text{ and }cos y = \sqrt{1-\frac{1}{10}}cosx=
1−
5
1
and cosy=
1−
10
1
cos x = \sqrt{\frac{4}{5}}\text{ and }cos y = \sqrt{\frac{9}{10}}cosx=
5
4
and cosy=
10
9
cos x =\frac{2}{ \sqrt{5}}\text{ and }cos y =\frac{3}{\sqrt{10}}cosx=
5
2
and cosy=
10
3
Now, we know that,
sin (x+y) = sin x × cos y + cos x × sin y
sin (x+y) = \frac{1}{\sqrt{5}}\times \frac{3}{\sqrt{10}} + \frac{2}{ \sqrt{5}}\times \frac{1}{\sqrt{10}}sin(x+y)=
5
1
×
10
3
+
5
2
×
10
1
sin (x+y) = \frac{3}{\sqrt{50}}+\frac{2}{ \sqrt{50}}sin(x+y)=
50
3
+
50
2
sin(x+y) = \frac{5}{\sqrt{50}}sin(x+y)=
50
5
sin(x+y) = \frac{5}{5\sqrt{2}}sin(x+y)=
5
2
5
sin(x+y) = \frac{1}{\sqrt{2}}sin(x+y)=
2
1
\implies (x+y) = sin^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}⟹(x+y)=sin
−1
(
2
1
)=
4
π
Hence, proved...