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Question: If \(\cos \theta + \sqrt 3 \sin \theta = 2\) then the value of \(\theta \)is equal to ?...

If cosθ+3sinθ=2\cos \theta + \sqrt 3 \sin \theta = 2 then the value of θ\theta is equal to ?

Explanation

Solution

we are give a trigonometric equation and with the known values sin30=12\sin {30^ \circ } = \dfrac{1}{2} and cos30=32\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}and the identity sin(A+B)=sinAcosB+cosAsinB\sin (A + B) = \sin A\cos B + \cos A\sin B we can find the value of θ\theta

Complete step by step solution:
We are given that cosθ+3sinθ=2\cos \theta + \sqrt 3 \sin \theta = 2
Diving by 2 on both sides we get,
12cosθ+32sinθ=1\Rightarrow \dfrac{1}{2}\cos \theta + \dfrac{{\sqrt 3 }}{2}\sin \theta = 1
We know that sin30=12\sin {30^ \circ } = \dfrac{1}{2} and cos30=32\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}
Substituting in the above equation we get
  sin30cosθ+sinθcos30=1\Rightarrow \;\sin {30^ \circ }\cos \theta + \sin \theta \cos {30^ \circ } = 1
By the identity
sin(A+B)=sinAcosB+cosAsinB\sin (A + B) = \sin A\cos B + \cos A\sin B
We get,
sin(30+θ)=1 30+θ=sin11 30+θ=90 θ=60  \Rightarrow \sin (30 + \theta ) = 1 \\\ \Rightarrow {30^ \circ } + \theta = {\sin ^{ - 1}}1 \\\ \Rightarrow {30^ \circ } + \theta = {90^ \circ } \\\ \Rightarrow \theta = {60^ \circ } \\\

Hence, the value of θ=60\theta = {60^ \circ }

Note:
There are six trigonometric functions: sine, cosine, tangent and their reciprocal functions, secant, cosecant and cotangent. These functions are found by the ratios of a triangle's sides
The trigonometric ratios for the angles 30°, 45° and 60° can be found using two special triangles.
An equilateral triangle with side lengths of 2 cm can be used to find exact values for the trigonometric ratios of 30° and 60°.