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Question

Question: Solve \(\cos \theta = \dfrac{1}{2}\)...

Solve cosθ=12\cos \theta = \dfrac{1}{2}

Explanation

Solution

We know that the two dimensional planes are divided into 4 quadrants as given below:

Quadrant I:0    π20\; - \;\dfrac{\pi }{2} All values are positive.
Quadrant II:π2    π\dfrac{\pi }{2}\; - \;\pi Only Sine and Cosec values are positive.
Quadrant III:π    3π2\pi \; - \;\dfrac{{3\pi }}{2} Only Tan and Cot values are positive.
Quadrant IV:3π2    2π\dfrac{{3\pi }}{2}\; - \;2\pi Only Cos and Sec values are positive.
Now using this basic knowledge we can find the quadrant where cosine would be positive.
Also in order to solve cosθ=12\cos \theta = \dfrac{1}{2}we have to find the value ofθ\theta .

Complete step by step solution:
Given
cosθ=12...........................(i)\cos \theta = \dfrac{1}{2}...........................\left( i \right)
Now we have to solve the equation (i), for that we have to find the value ofθ\theta . We know that the trigonometric values repeat after a certain interval and since cosine is a trigonometric function we have to find the intervals in which it repeats and thereby the different values of θ\theta .
On observing (i) we know can say from the standard formula that:
θ=π3........................(ii)\theta = \dfrac{\pi }{3}........................\left( {ii} \right)
Now we know that the cosine value is positive in Quadrant I and Quadrant IV so to find the next value we have to subtract the value given in (ii) with 2π2\pi to find the angle in the IV Quadrant.
i.e.

θ=2ππ3 θ=53π.......................(iii) \theta = 2\pi - \dfrac{\pi }{3} \\\ \theta = \dfrac{5}{3}\pi .......................\left( {iii} \right) \\\

Also now we know that the period in which cosine repeats is 2π2\pi which will repeat in both directions of2π2\pi .

Such that in general we can write:
θ=π3+2πnand5π3+2πn n:anyinteger  \theta = \dfrac{\pi }{3} + 2\pi n\,\,{\text{and}}\,\,\dfrac{{5\pi }}{3} + 2\pi n \\\ n:{\text{any}}\,{\text{integer}} \\\
Note: cosine values of some general angles are given below:

θ              cosθ π3                12 π6              32 π4              12 \theta \;\;\;\;\;\;\;\cos \theta \\\ \dfrac{\pi }{3}\;\;\;\;\;\;\;\;\dfrac{1}{2} \\\ \dfrac{\pi }{6}\;\;\;\;\;\;\;\dfrac{{\sqrt 3 }}{2} \\\ \dfrac{\pi }{4}\;\;\;\;\;\;\;\dfrac{1}{{\sqrt 2 }} \\\

While approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side.