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Question: If\[\sin A+\sin B+\sin C=3\], then\[\cos A+\cos B+\cos C\]is equal to\[(1)3\]\[(2)2\]\[(3)1\]\[(4)0\...

IfsinA+sinB+sinC=3\sin A+\sin B+\sin C=3, thencosA+cosB+cosC\cos A+\cos B+\cos Cis equal to(1)3$$$$(2)2$$$$(3)1$$$$(4)0

Explanation

Solution

Hint : To solve this question we will use the simple trigonometry. To get the solution of this question use the concept of range of sine function i.e. the maximum value attain by sine function is11 and value of sine and cosine function for different angles like00{{0}^{0}}and 900{{90}^{0}}.

Complete step-by-step solution:
Trigonometric functions are the functions that relate an angle of the right angled triangle to the ratios of two sides. The most used trigonometric functions in modern mathematics are sine, cosine and the tangent their abbreviation are sin, cos and tan. And their reciprocals are cosecant (cosec), secant (sec) and cotangent (cot) respectively.
In a right angled triangle, sine is the ratio of perpendicular side to the hypotenuse side whereas, cosine is the ratio of base side to the hypotenuse side and tangent is the ratio of perpendicular side to the hypotenuse side.
Trigonometry is the study of the relationship between angles and sides of the triangle. The angle varies003600{{0}^{0}}-{{360}^{0}}. The angles can be measured in degree as well as in radians.
We all know that the range of sine function is from 1to+1-1to+1i.e. [1,+1]\left[ -1,+1 \right]that means the highest value possible for the sine function is+1+1and the lowest value of the function is1-1. The same for the cosine function range is the same as of cosine function i.e. [1,+1]\left[ -1,+1 \right].
To get the solution of this question we will use the concept of this range value. In the question it is given that sinA+sinB+sinC=3\sin A+\sin B+\sin C=3and this situation is only possible when all of the three sines are equal to11. This means that
sinA=sinB=sinC=1\sin A=\sin B=\sin C=1
And we know that, if the sine value is 11 then, the angle between the sides is900{{90}^{0}}.
sin900+sin900+sin900=3\sin {{90}^{0}}+\sin {{90}^{0}}+\sin {{90}^{0}}=3
A=B=C=900A=B=C={{90}^{0}}
Now, substitute the value of A,BA,Band CCincosA+cosB+cosC\cos A+\cos B+\cos C, we get
cos900+cos900+cos900\cos {{90}^{0}}+\cos {{90}^{0}}+\cos {{90}^{0}}
And we know that the cos900=0\cos {{90}^{0}}=0
So, we get

& =0+0+0 \\\ & =0 \\\ \end{aligned}$$ Hence, we can conclude that$$\cos A+\cos B+\cos C=0$$. **So, the correct option is$$(4)$$.** **Note:** To calculate the values of the trigonometric functions we will always consider unit circle i.e. the circle of the radius equal to$$1$$ and the center point is$$(0,0)$$because this makes the calculation easy. In a unit circle there is a coordinates$$(\cos \theta ,\sin \theta )$$where$$\theta $$is the angle between $$x-axis$$and the unit radius in the anticlockwise direction.