# How to Simplify This Trigonometric Equation Using Substitutions?

• Fred1230
In summary: If so, you can use that to simplify the expression. In summary, the conversation is discussing a trigonometric identity involving the substitution method. The goal is to simplify the expression using trigonometric identities, specifically one for ##\cos(2\alpha)##.
Fred1230
Returning if I have to show the effort, I came to this:
$$\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.$$
=
$$\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.$$
I don't know how to use substitutions

$$s=\sin\alpha$$ and $$c=\cos\alpha$$

Fred1230 said:
Returning if I have to show the effort, I came to this:
$$\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.$$
=
$$\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.$$
I don't know how to use substitutions
Substitute ##\alpha=60^{\circ}## in your expression and check if you come out with ##\tan30^{\circ}##. If not it's back to the drawing board!

Fred1230 said:
Returning if I have to show the effort, I came to this:
$$\frac{\sin4\alpha}{1+\cos4\alpha}\cdot\frac{\cos2\alpha}{1+\cos2\alpha}\cdot\frac{\cos\alpha}{1+\cos\alpha}=\tan\frac{\alpha}{2}.$$
=
$$\frac{\sin4\alpha}{\sin^2\alpha+cos^2\alpha+\cos4\alpha}\cdot\frac{(\sin^2\alpha+cos^2\alpha)-2sin^2\alpha}{\sin^2\alpha+cos^2\alpha+\cos2\alpha}\cdot\frac{\cos\alpha}{\sin^2\alpha+cos^2\alpha+\cos\alpha}=\frac{\sin\alpha^2}{\cos2\alpha}.$$
I don't know how to use substitutions
Do you know a formula for ##\cos(2\alpha)## in terms of ##\cos(\alpha)##?

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