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Question: How do you solve \( \sin \theta = 1 \) ?...

How do you solve sin⁡θ=1\sin \theta = 1 ?

Explanation

Solution

In order to determine the value of the above question, first take inverse of sine on both of its side to pull out θ\theta from the inside of sine .Use the fact that the range and domain of the inverse of sine is in the interval [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] and x∈[−1,1].x \in \left[ { - 1,1} \right]. respectively to find the angle whose sine is equal to 1 to obtain your required result.

Complete step-by-step solution:
We are given a trigonometric function sin⁡θ=1\sin \theta = 1 .
Let’s take inverse of sine on both of its sides, we get
sin⁡−1(sin⁡θ)=sin⁡−1(1) θ=sin⁡−1(1)  {\sin ^{ - 1}}\left( {\sin \theta } \right) = {\sin ^{ - 1}}\left( 1 \right) \\\ \theta = {\sin ^{ - 1}}\left( 1 \right) \\\
We know that sin⁡−1θ{\sin ^{ - 1}}\theta denotes an angle in the interval [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] whose sine is xx for x∈[−1,1].x \in \left[ { - 1,1} \right].
Therefore,
sin⁡−1(1){\sin ^{ - 1}}\left( 1 \right) = An angle in [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] , whose sine is 11 .
From the trigonometric table we have,
sin⁡(π2)=1\sin \left( {\dfrac{\pi }{2}} \right) = 1
Transposing sin from left-hand side to right-hand side
sin⁡−1(1)=π2{\sin ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{2}

Hence, the required answer is π2\dfrac{\pi }{2} .

Note:
1. In Mathematics the inverse trigonometric functions (every so often additionally called anti-trigonometric functions or cyclomatic function) are the reverse elements of the mathematical functions In particular, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are utilized to get a point from any of the point's mathematical proportions. Reverse trigonometric functions are generally utilized in designing, route, material science, and calculation.
2. In inverse trigonometric function, the domain are the ranges of corresponding trigonometric functions and the range are the domain of the corresponding trigonometric function.
3. Periodic Function= A function f(x)f(x) is said to be a periodic function if there exists a real number T > 0 such that f(x+T)=f(x)f(x + T) = f(x) for all x.
If T is the smallest positive real number such that f(x+T)=f(x)f(x + T) = f(x) for all x, then T is called the fundamental period of f(x)f(x) .
Since sin⁡ (2nπ+θ)=sin⁡θ\sin \,(2n\pi + \theta ) = \sin \theta for all values of θ\theta and n ∈\in N.
4. Even Function – A function f(x)f(x) is said to be an even function ,if f(−x)=f(x)f( - x) = f(x) for all x in its domain.
Odd Function – A function f(x)f(x) is said to be an even function ,if f(−x)=−f(x)f( - x) = - f(x) for all x in its domain.
We know that sin⁡(−θ)=−sin⁡θ.cos⁡(−θ)=cos⁡θ and tan⁡(−θ)=−tan⁡θ\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta
Therefore, sin⁡θ\sin \theta and tan⁡θ\tan \theta and their reciprocals, csc⁡θ\csc\theta and cot⁡θ\cot \theta are odd functions whereas cos⁡θ\cos \theta and its reciprocal sec⁡θ\sec \theta are even functions.