Question

If x and y are acute angles such that sin x=1/√5 and sin y=1/√10 , prove that (x + y) = π/4

Answer 1

sin(x+y) = sin x * cos y +cos y * sin x = 1/√5 * 3/√(10) + (2/√5)*1/√(10) 
= (1/√2)*(3/5) + (1/√2)*(2/5) = (1/√2)*[(3/5)+(2/5)] =1/√2 
S0 x+y = π/4

Hope this helps!!!

Answer 2

Answer:

Given,

$sin x =\frac{1}{\sqrt{5}}\text{ and }sin y = \frac{1}{\sqrt{10}}$

$cos x = \sqrt{1-(\frac{1}{\sqrt{5}})^2}\text{ and }cos y = \sqrt{1-(\frac{1}{\sqrt{10}})^2}$

( Because sin²A + cos²A = 1 ⇒ cos²A = 1 -  sin²A ⇒ cos A = √(1-sin²A)

$cos x = \sqrt{1-\frac{1}{5}}\text{ and }cos y = \sqrt{1-\frac{1}{10}}$

$cos x = \sqrt{\frac{4}{5}}\text{ and }cos y = \sqrt{\frac{9}{10}}$

$cos x =\frac{2}{ \sqrt{5}}\text{ and }cos y =\frac{3}{\sqrt{10}}$

Now, we know that,

sin (x+y) = sin x × cos y + cos x × sin y

$sin (x+y) = \frac{1}{\sqrt{5}}\times \frac{3}{\sqrt{10}} + \frac{2}{ \sqrt{5}}\times \frac{1}{\sqrt{10}}$

$sin (x+y) = \frac{3}{\sqrt{50}}+\frac{2}{ \sqrt{50}}$

$sin(x+y) = \frac{5}{\sqrt{50}}$

$sin(x+y) = \frac{5}{5\sqrt{2}}$

$sin(x+y) = \frac{1}{\sqrt{2}}$

$\implies (x+y) = sin^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}$

Hence, proved...