Question

sin square 160 + sin square 140 + sin square 100 is equal to​

Answer 1

We know,

$\sin(180\textdegree-x)=\sin x\quad\implies\quad\sin^2(180\textdegree-x)=\sin^2x$

So we get,

$\begin{aligned}&\sin^2160\textdegree+\sin^2140\textdegree+\sin^2100\textdegree\\\\\implies\ \ &\sin^2(180\textdegree-20\textdegree)+\sin^2(180\textdegree-40\textdegree)+\sin^2(180\textdegree-80\textdegree)\\\\\implies\ \ &\sin^220\textdegree+\sin^240\textdegree+\sin^280\textdegree\\\\\implies\ \ &1-\cos^220\textdegree+1-\cos^240\textdegree+1-\cos^280\textdegree\\\\\implies\ \ &3-(\cos^220\textdegree+\cos^240\textdegree+\cos^280\textdegree)\end{aligned}$

But before, it is true that,

$\sin^220\textdegree+\sin^240\textdegree+\sin^280\textdegree=\cos^220\textdegree+\cos^240\textdegree+\cos^280\textdegree$

Because,

$\begin{aligned}&\sin^220\textdegree+\sin^240\textdegree+\sin^280\textdegree=\cos^220\textdegree+\cos^240\textdegree+\cos^280\textdegree\\\\\implies\ \ &\cos^220\textdegree+\cos^240\textdegree+\cos^280\textdegree-\sin^220\textdegree-\sin^240\textdegree-\sin^280\textdegree=0\\\\\implies\ \ &\cos^220\textdegree-\sin^220\textdegree+\cos^240\textdegree-\sin^240\textdegree+\cos^280\textdegree-\sin^280\textdegree=0\\\\\implies\ \ &\cos40\textdegree+\cos80\textdegree+\cos160\textdegree=0\end{aligned}$

$\begin{aligned}\implies\ \ &-\cos20\textdegree+\cos40\textdegree+\cos80\textdegree=0\\\\\implies\ \ &\cos40\textdegree+\cos80\textdegree=\cos20\textdegree\\\\\implies\ \ &2\cos\left(\dfrac{40\textdegree+80\textdegree}{2}\right)\cos\left(\dfrac{40\textdegree-80\textdegree}{2}\right)=\cos20\textdegree\\\\\implies\ \ &2\cos60\textdegree\cos20\textdegree=\cos20\textdegree\\\\\implies\ \ &2\cdot\dfrac{1}{2}\cos20\textdegree=\cos20\textdegree\end{aligned}$

So we can say that,

$\begin{aligned}&\sin^220\textdegree+\sin^240\textdegree+\sin^280\textdegree=\cos^220\textdegree+\cos^240\textdegree+\cos^280\textdegree\\\\\implies\ \ &3-(\cos^220\textdegree+\cos^240\textdegree+\cos^280\textdegree)=\cos^220\textdegree+\cos^240\textdegree+\cos^280\textdegree\\\\\implies\ \ &2(\cos^220\textdegree+\cos^240\textdegree+\cos^280\textdegree)=3\\\\\implies\ \ &\cos^220\textdegree+\cos^240\textdegree+\cos^280\textdegree=\mathbf{\dfrac{3}{2}}\end{aligned}$

Hence 3/2 is the answer.

Answer 2

Answer:

Here is the answer

3/2

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