Finding $2\sin x + 4\sin y$, given $\sin^2x+\cos^2y = \frac{11}{16}$ and $\sin\frac12(x+y) \cos\frac12(x-y) = \frac{5}{8}$

Question

I think my basics are pretty weak, I am not able to solve this question.

$$\begin{align} \sin^2x+\cos^2y &= \frac{11}{16} \tag1 \\[4pt] \sin\frac12(x+y) \cos\frac12(x-y) &= \frac{5}{8} \tag2\\ \\ \end{align}$$ Find the value of $$2\sin x + 4\sin y$$ (The answer is stated to be $4$.)

Sorry for the horrible $\LaTeX$.

MY APPROACH:

I did not focus on simplifying the first equation, so I simplified the second equation by using the formula $2\sin(x)\cos(y) = \sin(x + y) + \sin(x - y)$.

From 2

\begin{align*} \sin\left(\frac{x + y}{2}\right)\cos\left(\frac{x - y}{2}\right) & = \frac{5}{8}\\ 2\sin\left(\frac{x + y}{2}\right)\cos\left(\frac{x - y}{2}\right) & = \frac{5}{4}\\ \sin x + \sin y = \frac{5}{4} \end{align*}

Squaring this we get equation 3,

$$\sin^2x + \sin^2y + 2\sin x\sin y = \frac{25}{16}$$

Now I am adding equations 1 and 3 together

\begin{align*} \sin^2x + \sin^2y + 2\sin x \sin y + \sin^2x + \cos^2y & = \frac{25}{16} + \frac{11}{16}\\ \sin^2x + \sin^2y + 2\sin x \sin y + \sin^2x + \cos^2y & = \frac{36}{16}\\ \sin^2x + (\sin^2y + \cos^2y) + 2\sin x \sin y + \sin^2x & = \frac{9}{4}\\ 2\sin^2x + 2\sin x \sin y & = \frac{9}{4} \end{align*}

Taking $2\sin x$ as common

$$2\sin x(\sin x + \sin y) = \frac{9}{4}$$

But we know the value of $\boldsymbol{\sin x + \sin y}$ from equation number 2. Therefore,

\begin{align*} 2\sin x(\sin x + \sin y) & = \frac{9}{4}\\ 2\sin x \cdot \frac{5}{4} & = \frac{9}{4}\\ \sin x & = \frac{9}{10} \end{align*}

Then I put this value in equation 2 and got the value of $\sin y$ as $7/10$.

But when I used these values to find the solution for $2\sin x + 4\sin y$, I got a wrong answer $16/5$.

Answer 1

As you write in your post, $\;\color{brown }{\sin x+\sin y=\dfrac54}\,.$

Consequently, $\;\color{brown}{\sin x=\dfrac54-\sin y}\;.$

Moreover, from $\;\color{brown}{\sin^2x+\cos^2y=\dfrac{11}{16}}\;,\;$ it follows that

$\left(\dfrac54-\sin y\right)^2+1-\sin^2y=\dfrac{11}{16}\;,$

$\dfrac{25}{16}+\sin^2y-\dfrac52\sin y+1-\sin^2y=\dfrac{11}{16}\;,$

$\sin y=\dfrac34\,.$

Consequently,

$\begin{align}2\sin x+4\sin y&=2\left(\sin x+\sin y\right)+2\sin y=\\[3pt]&=2\cdot\dfrac54+2\cdot\dfrac34=4\,.\end{align}$

Answer 2

As you calculated, $\sin(x) + \sin(y) = 5/4$. therefore, $2\sin(x) + 2\sin(y) = 2.5$ . The question has required you to calculate $ 2\sin(x) + 4\sin(y)$ which is 2.5 + 2 sin(y). In order to calculate sin(y), put $\sin(x) = 5/4 - \sin(y)$ and use the other equation $ \sin^2x + \cos^2y = 11/16$ and put $\cos^2y = 1 - \sin^2y$ to solve the exercise. I calculated it and got 4.