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Question

Question: Express \[\sin \dfrac{x}{2}\] in terms of \(\cos x\) using the double angle identity?...

Express sinx2\sin \dfrac{x}{2} in terms of cosx\cos x using the double angle identity?

Explanation

Solution

We use the formula for the trigonometric function of multiple angles where cosx=12sin2x2\cos x=1-2{{\sin }^{2}}\dfrac{x}{2}. We simplify the equation to get the value of sinx2\sin \dfrac{x}{2} in terms of cosx\cos x. The final expression is the solution of the problem.

Complete step by step answer:
We have the formula for the trigonometric function of submultiple angles where cosx=12sin2x2\cos x=1-2{{\sin }^{2}}\dfrac{x}{2}. We need to find the inverse form of cosx=12sin2x2\cos x=1-2{{\sin }^{2}}\dfrac{x}{2} to get the value of sinx2\sin \dfrac{x}{2} in terms of cosx\cos x.
Simplifying the equation, we get
cosx=12sin2x2 2sin2x2=1cosx \cos x=1-2{{\sin }^{2}}\dfrac{x}{2} \\\ \Rightarrow 2{{\sin }^{2}}\dfrac{x}{2}=1-\cos x \\\
We now divide both sides by 2 to get
2sin2x2=1cosx sin2x2=1cosx2 2{{\sin }^{2}}\dfrac{x}{2}=1-\cos x \\\ \Rightarrow {{\sin }^{2}}\dfrac{x}{2}=\dfrac{1-\cos x}{2} \\\
Now we take the square root value of the equation and get sin2x2=±1cosx2\sqrt{{{\sin }^{2}}\dfrac{x}{2}}=\pm \sqrt{\dfrac{1-\cos x}{2}}.
Expressing sinx2\sin \dfrac{x}{2} in terms of cosx\cos x, we get sinx2=±1cosx2\sin \dfrac{x}{2}=\pm \sqrt{\dfrac{1-\cos x}{2}}.

Hence, the simplified form will be sinx2=±1cosx2\sin \dfrac{x}{2}=\pm \sqrt{\dfrac{1-\cos x}{2}}.

Note: The multiple angles work as cos2x=12sin2x\cos 2x=1-2{{\sin }^{2}}x. We just change the angle from 2x2x to xx to get cosx=12sin2x2\cos x=1-2{{\sin }^{2}}\dfrac{x}{2}. The trigonometric functions of multiple angles are the multiple angle formula. Double and triple angles formulas are there under the multiple angle formulas. Sine, tangent and cosine are the general functions for the multiple angle formula.