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Question: How do you solve \(\cos \theta =\sin \theta \) ?...

How do you solve cos⁡θ=sin⁡θ\cos \theta =\sin \theta ?

Explanation

Solution

In this question we will convert the trigonometric identity using trigonometric equivalences we will try to solve the range of values for the angle θ\theta for which the value of sin⁡θ\sin \theta and cos⁡θ\cos \theta is same and then simplify the expression to get the required value of θ\theta and write the solution in the general format which is the final answer.

Complete step-by-step answer:
We have the given equation as:
⇒cos⁡θ=sin⁡θ\Rightarrow \cos \theta =\sin \theta
Now we know that the value of sin⁡θ\sin \theta and cos⁡θ\cos \theta toggles after every π2\dfrac{\pi }{2} radians.
Therefore, we can write sin⁡θ\sin \theta in the form of cos⁡θ\cos \theta as:
⇒cos⁡θ=cos⁡(±π2−θ)\Rightarrow \cos \theta =\cos \left( \pm \dfrac{\pi }{2}-\theta \right)
Now since both the trigonometric functions are the same on both the sides of the equation, we can remove them. On removing it is to be remembered that the solution will toggle after even 2π2\pi radians.
Therefore, we can write the angles as:
⇒θ=±π2−θ+2kπ\Rightarrow \theta =\pm \dfrac{\pi }{2}-\theta +2k\pi , where kkis any integer.
Now on taking θ\theta on the same side of the expression, we get:
⇒θ+θ=±π2+2kπ\Rightarrow \theta +\theta =\pm \dfrac{\pi }{2}+2k\pi
On simplifying the left-hand side of the expression, we get:
⇒2θ=±π2+2kπ\Rightarrow 2\theta =\pm \dfrac{\pi }{2}+2k\pi
Now on dividing both the sides of the expression by 22, we get:
⇒θ=±π4+kπ\Rightarrow \theta =\pm \dfrac{\pi }{4}+k\pi , which is the required solution.

Note: Now to check whether the solution, the above angle should be substituted in cos⁡θ\cos \theta and sin⁡θ\sin \theta and evaluated. If both the values are the same, the value of θ\theta calculated is correct.
For simplicity purposes, we will consider k=1k=1 and the solution in the first quadrant which implies that we are considering θ\theta as positive.
Consider cos⁡(π4+1×π)\cos \left( \dfrac{\pi }{4}+1\times \pi \right)
On simplifying, we get:
⇒cos⁡(5π4)\Rightarrow \cos \left( \dfrac{5\pi }{4} \right)
Which has a value using calculator as: −12-\dfrac{1}{\sqrt{2}}
Now consider sin⁡(π4+1×π)\sin \left( \dfrac{\pi }{4}+1\times \pi \right)
On simplifying, we get:
⇒sin⁡(5π4)\Rightarrow \sin \left( \dfrac{5\pi }{4} \right)
Which has a value using calculator as: −12-\dfrac{1}{\sqrt{2}}
Therefore, we can conclude that cos⁡θ=sin⁡θ\cos \theta =\sin \theta , therefore the solution is correct.
The trigonometric table should be remembered while doing these types of questions.
The principal values of sin⁡\sin and cos⁡\cos should be remembered.