Question

The value of \[\theta \] satisfying \[\cos \theta +\sqrt{3}\sin \theta =2\] is:
A. \[{{30}^{\circ }}\]
B. \[{{60}^{\circ }}\]
C. \[{{45}^{\circ }}\]
D. \[{{90}^{\circ }}\]

Answer 1

Hint:First of all we will divide the given equation by \[2\] on both sides. After that put the values of trigonometric ratios of standard angles then apply the formula of difference of two angles to solve the rest of the equation then check which option is correct among the above given options.

Complete step by step answer:
The word Trigonometry is derived from the Greek word trigon and metron. Here trigon means figures with three angles and metron means measurement. That is the meaning of the word Trigonometry is measurement of triangles. In the modern age Trigonometry has broad based meaning. In simple words we define it as that branch of mathematics which deals with the measurement of angles and the problems allied with angles.
Currently trigonometry is used in many areas such as the science of seismology, designing electric circuits, predicting the heights of tides in the ocean, analyzing the musical tone and in many other areas.
We know that the angle between \[{{0}^{\circ }}\] and \[{{90}^{\circ }}\] is called acute angle. Further the angle of \[{{90}^{\circ }}\] is called right angle and the angle lies between \[{{90}^{\circ }}\] and \[{{180}^{\circ }}\] is called obtuse angle.
A series of real valued functions, defined as the ratio of the sides of a triangle is called a Trigonometric Function.
Trigonometric functions are also known as the circular functions.The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. The angles of sine, cosine, and tangent are the primary classification of functions of trigonometry. And the three functions which are cotangent, secant and cosecant can be derived from the primary functions.

We have given that the equation is \[\cos \theta +\sqrt{3}\sin \theta =2\]
Dividing the whole equation by \[2\] we will get:
\[\Rightarrow \]\[\dfrac{1}{2}\cos \theta +\dfrac{1}{2}\sqrt{3}\sin \theta =\dfrac{2}{2}\]
\[\Rightarrow \]\[\dfrac{1}{2}\cos \theta +\dfrac{1}{2}\sqrt{3}\sin \theta =1\]
As we know that \[\cos {{60}^{\circ }}=\dfrac{1}{2}\] and \[\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\] putting these values in the equation we will get:
\[\Rightarrow \]\[\cos {{60}^{\circ }}\cos \theta +\sin {{60}^{\circ }}\sin \theta =1\]
Now according to the formula \[\cos A\cos B+\sin A\sin B=\cos (A-B)\] , where \[A={{60}^{\circ }}\] and \[B =\theta \]
So, we will get
\[\Rightarrow \]\[\cos ({{60}^{\circ }}-\theta )=1\]
And we know that \[\cos {{0}^{\circ }}=1\]
\[\Rightarrow \]\[\cos ({{60}^{\circ }}-\theta )=\cos {{0}^{\circ }}\]
\[\Rightarrow \]\[{{60}^{\circ }}-\theta ={{0}^{\circ }}\]
Hence the value of \[\theta ={{60}^{\circ }}\]

So, the correct answer is “Option B”.

Note: We must keep one thing in mind that \[\sin \theta \] is not the same as \[\sin \times \theta \] because it represents a ratio, not a product and this is true for all the trigonometric ratios. Any trigonometric function of angle \[{{\theta }^{\circ }}\] is equal to the same trigonometric function of any angle \[n\times {{360}^{\circ }}+\theta \], where \[n\] is any integer.