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Question: If \(\alpha + \beta + \gamma = \pi \), then \({\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma \) ...

If α+β+γ=π\alpha + \beta + \gamma = \pi , then sin⁡2α+sin⁡2β−sin⁡2γ{\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma is equal to
A.2sin⁡αsin⁡βcos⁡γ2\sin \alpha \sin \beta \cos \gamma
B.2cos⁡αcos⁡βcos⁡γ2\cos \alpha \cos \beta \cos \gamma
C.2sin⁡αsin⁡βsin⁡γ2\sin \alpha \sin \beta \sin \gamma
D.None of the above

Explanation

Solution

In order to solve the given question, we will simplify the given angle in terms of α\alpha and β\beta . Then we will put some fundamental formulas of trigonometry. After all of that we will do further simplification and put our angle values and find out our final answers. Some special formulas are used in this question are,
sin⁡2A=1−cos⁡2A{\sin ^2}A = 1 - {\cos ^2}A
sin⁡2A=1−cos⁡2A2{\sin ^2}A = \dfrac{{1 - \cos 2A}}{2}

Complete answer:
First of all, we have given α+β+γ=π\alpha + \beta + \gamma = \pi , we can write this term also in this way α+β=π−γ\alpha + \beta = \pi - \gamma . We can multiply it by 2 and we get 2α+2β=2π−2γ2\alpha + 2\beta = 2\pi - 2\gamma .
Now, Our given question is
⇒sin⁡2α+sin⁡2β−sin⁡2γ\Rightarrow {\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma
Now, replace some terms with following formula,
sin⁡2A=1−cos⁡2A{\sin ^2}A = 1 - {\cos ^2}A
And
sin⁡2A=1−cos⁡2A2{\sin ^2}A = \dfrac{{1 - \cos 2A}}{2}.
After replacement we get,
⇒(1−cos⁡2α2)+(1−cos⁡2β2)−(1−cos⁡2γ)\Rightarrow (\dfrac{{1 - \cos 2\alpha }}{2}) + (\dfrac{{1 - \cos 2\beta }}{2}) - (1 - {\cos ^2}\gamma )
Take −(12) - (\dfrac{1}{2}) common and implement this formula cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})
⇒1−12(cos⁡2α+cos⁡2β)−(1−cos⁡2γ)\Rightarrow 1 - \dfrac{1}{2}(\cos 2\alpha + \cos 2\beta ) - (1 - {\cos ^2}\gamma )
After some for simplification, we get
⇒−12(2cos⁡(α+β)cos⁡(α−β))+cos⁡2γ\Rightarrow - \dfrac{1}{2}(2\cos (\alpha + \beta )\cos (\alpha - \beta )) + {\cos ^2}\gamma
But as we know α+β=π−γ\alpha + \beta = \pi - \gamma and cos⁡(π+A)=−cos⁡A\cos (\pi + A) = - \cos A so we can do further steps,
⇒−cos⁡(π+γ)cos⁡(α−β)+cos⁡2γ\Rightarrow - \cos (\pi + \gamma )\cos (\alpha - \beta ) + {\cos ^2}\gamma
On further simplification we get,
⇒cos⁡γcos⁡(α−β)+cos⁡2γ\Rightarrow \cos \gamma \cos (\alpha - \beta ) + {\cos ^2}\gamma
Now, let’s come to our final step. In this step we will take cos⁡γ\cos \gamma common and implement cos⁡A−cos⁡B=2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = 2\sin (\dfrac{{A + B}}{2})\sin (\dfrac{{A - B}}{2}) this formula we will get our answer.
⇒cos⁡γ(cos⁡(α−β)+cos⁡(α+β−π)\Rightarrow \cos \gamma (\cos (\alpha - \beta ) + \cos (\alpha + \beta - \pi )
we know that (cos⁡(−A)=cos⁡A\cos ( - A) = \cos A)
⇒cos⁡γ(cos⁡(α−β)+cos⁡(π−(α+β))\Rightarrow \cos \gamma (\cos (\alpha - \beta ) + \cos (\pi - (\alpha + \beta ))
Do some more calculation and we get,
⇒cos⁡γ(cos⁡(α−β)−cos⁡(α+β))\Rightarrow \cos \gamma (\cos (\alpha - \beta ) - \cos (\alpha + \beta ))
⇒cos⁡γ(2sin⁡(α−β+α+β2)sin⁡(α+β−(α−β)2))\Rightarrow \cos \gamma (2\sin (\dfrac{{\alpha - \beta + \alpha + \beta }}{2})\sin (\dfrac{{\alpha + \beta - (\alpha - \beta )}}{2}))
Just cancel out bracket things and we get,
⇒2sin⁡αsin⁡βcos⁡γ\Rightarrow 2\sin \alpha \sin \beta \cos \gamma.
Hence, sin⁡2α+sin⁡2β−sin⁡2γ = 2sin⁡αsin⁡βcos⁡γ{\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma \, = \,2\sin \alpha \sin \beta \cos \gamma .
Therefore the correct answer is option A .

Note:
The sum of the squares of sine and cosine angle is equal to 1. It is an identity so it can be used everywhere or to solve these types of problems but there are some certain formulae that are valid at some definite interval. So in order to solve these types of problems one should check the formulae and their intervals where they hold true.