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Question: How do you simplify \({{\sin }^{2}}\theta \) to non-exponential trigonometric functions?...

How do you simplify sin⁡2θ{{\sin }^{2}}\theta to non-exponential trigonometric functions?

Explanation

Solution

We have to convert sin⁡2θ{{\sin }^{2}}\theta into non-exponential trigonometric functions. For this we will use the trigonometric identities and formulas. We will use the following identity and relation to solve this question:
cos⁡2θ=1−2sin⁡2θ\cos 2\theta =1-2{{\sin }^{2}}\theta
sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1

Complete step by step answer:
We have been given a trigonometric function sin⁡2θ{{\sin }^{2}}\theta .
We have to simplify the given function and convert it into a non-exponential trigonometric function.
Now, we know that trigonometry is a branch of mathematics that deals with the study of measuring the sides and angles of a triangle, mainly the right triangle. Trigonometric identities can be used to solve trigonometric problems easily by reducing the number of computational steps.
Now, we know that there are some trigonometric identities which establish a relation between trigonometric ratios.
Converting the given function into non-exponential form means we have to reduce the power of the function to 1.
We know that cos⁡2θ=1−2sin⁡2θ\cos 2\theta =1-2{{\sin }^{2}}\theta
We can rewrite the above identity as
⇒2sin⁡2θ=1−cos⁡2θ ⇒sin⁡2θ=1−cos⁡2θ2 \begin{aligned} & \Rightarrow 2{{\sin }^{2}}\theta =1-\cos 2\theta \\\ & \Rightarrow {{\sin }^{2}}\theta =\dfrac{1-\cos 2\theta }{2} \\\ \end{aligned}

Hence we get the non-exponential form of sin⁡2θ{{\sin }^{2}}\theta as 1−cos⁡2θ2\dfrac{1-\cos 2\theta }{2}.

Note: As there are six trigonometric ratios sine, cosine, tangent, cotangent, secant and cosecant. Students must have the knowledge of some basic trigonometric identities to solve this type of question. Trigonometric identities establish a relation between these trigonometric ratios. We can easily convert a ratio into another form by using these identities. Alternatively we can also use other identities of trigonometry to solve the given question such as:
cos⁡2θ=cos⁡2θ−sin⁡2θ cos⁡2θ=2cos⁡2θ−1 \begin{aligned} & \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\\ & \cos 2\theta =2{{\cos }^{2}}\theta -1 \\\ \end{aligned}