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Question: If \(\sin \alpha + \sin \beta + \sin \gamma = 0 = \cos \alpha + \cos \beta + \cos \gamma \) then \({...

If sin⁡α+sin⁡β+sin⁡γ=0=cos⁡α+cos⁡β+cos⁡γ\sin \alpha + \sin \beta + \sin \gamma = 0 = \cos \alpha + \cos \beta + \cos \gamma then sin⁡2α+sin⁡2β+sin⁡2γ={\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma =
A.−32 - \dfrac{3}{2}
B.32\dfrac{3}{2}
C.23\dfrac{2}{3}
D.None of these

Explanation

Solution

Here we will use some trigonometric formula for step by step and we will solve the answer for this question. Then we find which answer is correct for the given option. Here we will use a double angle formula. Here we will use a double angle formula. It means angles are doubled values.

Formula used:
cos⁡2a=1−sin⁡2a\cos 2a = 1 - {\sin ^2}a

Complete step-by-step answer:
Given Question is
sin⁡α+sin⁡β+sin⁡γ=0=cos⁡α+cos⁡β+cos⁡γ\sin \alpha + \sin \beta + \sin \gamma = 0 = \cos \alpha + \cos \beta + \cos \gamma
Let a=cos⁡α+isin⁡α,b=cos⁡β+isin⁡βa = \cos \alpha + i\sin \alpha ,b = \cos \beta + i\sin \beta and c=cos⁡γ+isin⁡γc = \cos \gamma + i\sin \gamma
Here a means we will multiply sin⁡α\sin \alpha and cos⁡α\cos \alpha we will store the answer and again multiply two and store the values bb and c
a+b+c=0a + b + c = 0 and 1a+1b+1c=0\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 0 (we will plus a,ba,b and ccvalues in the equation)
Here we will take above equation
⇒a2+b2+c2=0\Rightarrow {a^2} + {b^2} + {c^2} = 0
We substitute the above values we will get the answer
∴cos⁡2α+cos⁡2β+cos⁡2γ=0\therefore \cos 2\alpha + \cos 2\beta + \cos 2\gamma = 0
Using formula cos⁡2a=1−sin⁡2a\cos 2a = 1 - {\sin ^2}a for above equation we will get
⇒1−2sin⁡2α+1−2sin⁡2β+1−2sin⁡2γ=0\Rightarrow 1 - 2{\sin ^2}\alpha + 1 - 2{\sin ^2}\beta + 1 - 2{\sin ^2}\gamma = 0
We will simplify the above equation we will get the answer for the given question
∴sin⁡2α+sin⁡2β+sin⁡2γ=32\therefore {\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma = \dfrac{3}{2}
Hence, the correct option is B.

Additional information:
Here we have seen special cases of the sum and difference formulas for sine and cosine yields what is known as the double‐angle identities and the half‐angle identities. Many functions involving powers of sine and cosine are hard to integrate. The use of Double-Angle formulas help reduce the degree of difficulty. An expression involving the trigonometric functions with their first power.

Note: Here we will concentrate square and add the value in above solution. Given the trigonometric values of an angle α\alpha , we would be able to determine the trigonometric values for another angle 2α2\alpha . Here also we will concentrate the angle β\beta and the angle γ\gamma for the question that will be important.